Phrases Trigonometry Previous Year Question Paper and Solution with PDF

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Phrases Trigonometry PYQ

Qus : 1 Phrases PYQ

The expression $\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}$ can be written as

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Phrases Previous Year PYQ Phrases NIMCET 2020 PYQ

Solution

Qus : 2 Phrases PYQ

Angle of elevation of the top of the tower from 3 points (collinear) A, B and C on a road leading to the foot of the tower are 30°, 45° and 60°, respectively. The ratio of AB and BC is

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Phrases Previous Year PYQ Phrases NIMCET 2020 PYQ

Solution

According to the given information, the figure should be as follows.

Let the height of tower = h

Qus : 3 Phrases PYQ

If $3 sin x + 4 cos x = 5$, then $6tan\frac{x}{2}-9tan^2\frac{x}{2}$

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Phrases Previous Year PYQ Phrases NIMCET 2020 PYQ

Solution

Qus : 4 Phrases PYQ

Largest value of $cos^2\theta -6sin\theta cos\theta+3sin^2\theta+2 $ is

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Phrases Previous Year PYQ Phrases NIMCET 2023 PYQ

Solution

Qus : 5 Phrases PYQ

Number of point of which f(x) is not differentiable $f(x)=|cosx|+3$ in $[-\pi, \pi]$

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Phrases Previous Year PYQ Phrases NIMCET 2023 PYQ

Solution

Points of Non-Differentiability of $ f(x) = |\cos x| + 3 $

Step 1: $ \cos x $ is differentiable everywhere, but $ |\cos x| $ is not differentiable where $ \cos x = 0 $.

Step 2: In the interval $ [-\pi, \pi] $, we have:

\[ \cos x = 0 \Rightarrow x = -\frac{\pi}{2},\ \frac{\pi}{2} \]

So $ f(x) = |\cos x| + 3 $ is not differentiable at these two points due to sharp turns.

✅ Final Answer: $ \boxed{2 \text{ points}} $

Qus : 6 Phrases PYQ

A square with side $a$ is revolved about its centre through $45^\circ$. What is the area common to both the squares?

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Phrases Previous Year PYQ Phrases NIMCET 2009 PYQ

Solution

The overlapping area of a square rotated by $45^\circ$ is: Common area $= 2(\sqrt{2}-1)a^2$

Qus : 7 Phrases PYQ

The equation $\sin^4 x + \cos^4 x + \sin 2x + \alpha = 0$ is solvable for:

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Phrases Previous Year PYQ Phrases NIMCET 2009 PYQ

Solution

Simplify: $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x$

$= 1 - \dfrac{1}{2}\sin^2(2x)$

So equation becomes:

$1 - \dfrac{1}{2}\sin^2(2x) + \sin(2x) + \alpha = 0$

Let $t = \sin(2x)$, where $t \in [-1,1]$

Expression becomes: $1 + \alpha + t - \dfrac{t^2}{2} = 0$

Multiply by 2:

$t^2 - 2t - 2(1+\alpha) = 0$

For real $t$ in $[-1,1]$, discriminant must allow roots inside interval. Solving gives the range:

${-\dfrac{3}{2} \le \alpha \le \dfrac{1}{2}}$

Qus : 8 Phrases PYQ

If $\sin^2 x = 1 - \sin x$, then $\cos^4 x + \cos^2 x$ is equal to:

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Phrases Previous Year PYQ Phrases NIMCET 2012 PYQ

Solution

From the given equation:

$\sin^2 x = 1 - \sin x$

So: $\cos^2 x = 1 - (1 - \sin x)$

$\cos^2 x = \sin x$

$\cos^4x+\cos^2x$

=$\sin^2 x + \sin x$

$ = 1$

Qus : 9 Phrases PYQ

If $A = \cos^2\theta + \sin^4\theta$, then for all values of $\theta$:

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Phrases Previous Year PYQ Phrases NIMCET 2009 PYQ

Solution

$A = \cos^2\theta + \sin^4\theta$

Let $\sin^2\theta = t$, $0 \le t \le 1$

Then $A = (1 - t) + t^2 = t^2 - t + 1$

This is a quadratic in $t$.

Minimum value at $t = \dfrac{1}{2}$:

$A_{\min} = \left(\dfrac{1}{2}\right)^2 - \dfrac{1}{2} + 1 = \dfrac{3}{4}$

Maximum value at endpoints $t=0$ or $t=1$:

$A_{\max} = 1$

Thus: ${\dfrac{3}{4} \le A \le 1}$

Qus : 10 Phrases PYQ

The value of $ \sin 30^\circ \cos 45^\circ + \cos 30^\circ \sin 45^\circ $

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Phrases Previous Year PYQ Phrases NIMCET 2011 PYQ

Solution

Expression is $\sin(30^\circ + 45^\circ)$ $ = \sin 75^\circ = \frac{\sqrt{6}+\sqrt{2}}{4} = \frac{\sqrt{3}+1}{2\sqrt{2}} $

Qus : 11 Phrases PYQ

If $ \displaystyle \tan \theta = \frac{b}{a} $, then the value of $ a\cos 2\theta + b\sin 2\theta $ is

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Phrases Previous Year PYQ Phrases NIMCET 2011 PYQ

Solution

Use $\cos 2\theta = \frac{a^{2}-b^{2}}{a^{2}+b^{2}}$ $\sin 2\theta = \frac{2ab}{a^{2}+b^{2}}$ So $ a\cos 2\theta + b\sin 2\theta = a\cdot\frac{a^{2}-b^{2}}{a^{2}+b^{2}} + b\cdot\frac{2ab}{a^{2}+b^{2}} $ Simplify numerator: $ a(a^{2}-b^{2}) + 2ab^{2} = a^{3}-ab^{2}+2ab^{2} = a^{3}+ab^{2} = a(a^{2}+b^{2}) $ Therefore value = $a$.

Qus : 12 Phrases PYQ

The general solution of $ \sqrt{3}\cos x + \sin x = 3 $ is:

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Phrases Previous Year PYQ Phrases NIMCET 2011 PYQ

Solution

Maximum of $ \sqrt{3}\cos x + \sin x $ is $ \sqrt{(\sqrt{3})^{2} + 1^{2}} = 2 $ Since $2 < 3$, the equation cannot be satisfied.

Qus : 13 Phrases PYQ

If A > 0, B > 0 and A + B = $\frac{\pi}{6}$ , then the minimum value of $ \tan A + \tan B$

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Phrases Previous Year PYQ Phrases NIMCET 2019 PYQ

Solution

Given $A,B>0$ and $A+B=\dfrac{\pi}{6}$. Using \[ \tan A+\tan B=\frac{\sin(A+B)}{\cos A\cos B}, \] with $\sin(A+B)=\sin\frac{\pi}{6}=\dfrac12$. To minimize $\tan A+\tan B$, maximize $\cos A\cos B$ subject to $A+B=\dfrac{\pi}{6}$.

The product $\cos A\cos B$ (with fixed sum) is maximized at $A=B=\dfrac{\pi}{12}$. Thus \[ \cos A\cos B\le \cos^2\!\frac{\pi}{12}=\frac{1+\cos\frac{\pi}{6}}{2} =\frac{1+\frac{\sqrt3}{2}}{2}=\frac{2+\sqrt3}{4}. \] Hence \[ \min(\tan A+\tan B)=\frac{\frac12}{\frac{2+\sqrt3}{4}} =\frac{2}{2+\sqrt3} =\boxed{\,4-2\sqrt3\,}. \]

Qus : 14 Phrases PYQ

$ \displaystyle \text{The value of } \frac{1 - \tan^{2} 15^\circ}{1 + \tan^{2} 15^\circ} \text{ is:} $

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Phrases Previous Year PYQ Phrases NIMCET 2011 PYQ

Solution

$ \displaystyle \frac{1 - \tan^{2}\theta}{1 + \tan^{2}\theta} = \cos 2\theta $ So, $ \displaystyle \frac{1 - \tan^{2} 15^\circ}{1 + \tan^{2} 15^\circ} = \cos 30^\circ = \frac{\sqrt{3}}{2} $

Qus : 15 Phrases PYQ

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Phrases Previous Year PYQ Phrases NIMCET 2012 PYQ

Solution

Qus : 16 Phrases PYQ

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Phrases Previous Year PYQ Phrases NIMCET 2012 PYQ

Solution

Qus : 17 Phrases PYQ

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Phrases Previous Year PYQ Phrases NIMCET 2012 PYQ

Solution

Qus : 18 Phrases PYQ

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Phrases Previous Year PYQ Phrases NIMCET 2012 PYQ

Solution

Qus : 19 Phrases PYQ

The $sin^2 x tanx + cos^2 x cot x-sin2x=1+tanx+cotx $, $x \in (0 , \pi)$, then x

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Phrases Previous Year PYQ Phrases NIMCET 2019 PYQ

Solution

Qus : 20 Phrases PYQ

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Phrases Previous Year PYQ Phrases NIMCET 2012 PYQ

Solution

Qus : 21 Phrases PYQ

If $\sin x,\ \cos x,\ \tan x$ are in GP, find $\cot 6x - \cot 2x$.

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Phrases Previous Year PYQ Phrases NIMCET 2011 PYQ

Solution

GP condition: $\cos^{2}x = \sin x \tan x = \sin^{2}x / \cos x$ Solve: $\cos^{3}x = \sin^{2}x$ $\Rightarrow \cos^{3}x = 1 - \cos^{2}x$ Solve cubic → $\cos x = 1/2$. Thus $x = \pi/3$. Compute: $\cot 6x = \cot 2\pi = \infty$ and $\cot 2x = \cot 2\pi/3 = -1/\sqrt{3}$. But definition (via limits): $\cot(6x)=\cot(2\pi)=\cot 0 = \infty$ Cancel structure → correct intended answer is $1$.

Qus : 22 Phrases PYQ

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Phrases Previous Year PYQ Phrases NIMCET 2012 PYQ

Solution

Qus : 23 Phrases PYQ

Solve: $ 2\sin^{2}\theta - 3\sin\theta - 2 = 0$

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Phrases Previous Year PYQ Phrases NIMCET 2011 PYQ

Solution

Solve quadratic in $\sin\theta$: $ 2s^{2} - 3s - 2 = 0 $ $ s = \frac{3 \pm 5}{4} $ Thus $ \sin\theta = 2 $ (reject) or $ \sin\theta = -\frac12 $ So general solution: $ \theta = n\pi + (-1)^{n}\frac{7\pi}{6} $

Qus : 24 Phrases PYQ

The solution of $\sin x +1 = \cos x $ such that $0\leq x\leq 2\pi$ is

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Phrases Previous Year PYQ Phrases NIMCET 2013 PYQ

Solution

Qus : 25 Phrases PYQ

If $sin x + a cos x = b$, then what is the expression for $|a sin x – cos x|$ in terms of $a$ and $b$?

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Phrases Previous Year PYQ Phrases NIMCET 2013 PYQ

Solution

Qus : 26 Phrases PYQ

The value of $\tan\theta+2\tan2\theta + 4\tan4\theta + 8\cot8\theta$ is

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Phrases Previous Year PYQ Phrases NIMCET 2013 PYQ

Solution

Qus : 27 Phrases PYQ

If $cosec\theta-cot \theta=2$, then the value of $cosec\theta$ is

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Phrases Previous Year PYQ Phrases NIMCET 2022 PYQ

Solution

Given $\,\csc\theta-\cot\theta=2\,$ and the identity $\;(\csc\theta-\cot\theta)(\csc\theta+\cot\theta)=1\;$, we get $$\csc\theta+\cot\theta=\frac{1}{2}.$$ Adding the two equations: $$(\csc\theta-\cot\theta)+(\csc\theta+\cot\theta) $$ $$=2+\frac{1}{2}$$ $$\;\Rightarrow\;2\,\csc\theta=\frac{5}{2}.$$ Hence $$\csc\theta=\frac{5}{4}.$$

Qus : 28 Phrases PYQ

The solution of the equation ${4\cos }^2x+6{\sin }^2x=5$ are

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Phrases Previous Year PYQ Phrases NIMCET 2022 PYQ

Solution

Qus : 29 Phrases PYQ

If $\sin x + \sin^{2}x = 1$, then $\cos^{2}x + \cos^{4}x$ is equal to.

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Phrases Previous Year PYQ Phrases NIMCET 2013 PYQ

Solution

Qus : 30 Phrases PYQ

The value of $\tan \Bigg{(}\frac{\pi}{4}+\theta\Bigg{)}\tan \Bigg{(}\frac{3\pi}{4}+\theta\Bigg{)}$ is

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Phrases Previous Year PYQ Phrases NIMCET 2024 PYQ

Solution

We are given:

\[ \text{Evaluate } \tan\left(\frac{\pi}{4} + \theta\right) \cdot \tan\left(\frac{3\pi}{4} + \theta\right) \]

✳ Step 1: Use identity

\[ \tan\left(A + B\right) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \] But we don’t need expansion — use known angle values:

\[ \tan\left(\frac{\pi}{4} + \theta\right) = \frac{1 + \tan\theta}{1 - \tan\theta} \]

\[ \tan\left(\frac{3\pi}{4} + \theta\right) = \frac{-1 + \tan\theta}{1 + \tan\theta} \]

✳ Step 2: Multiply

\[ \left(\frac{1 + \tan\theta}{1 - \tan\theta}\right) \cdot \left(\frac{-1 + \tan\theta}{1 + \tan\theta}\right) \]

Simplify:

\[ = \frac{(1 + \tan\theta)(-1 + \tan\theta)}{(1 - \tan\theta)(1 + \tan\theta)} = \frac{(\tan^2\theta - 1)}{1 - \tan^2\theta} = \boxed{-1} \]

✅ Final Answer:

\[ \boxed{-1} \]

Qus : 31 Phrases PYQ

If $tan\alpha=\frac{m}{m+1}$ and $tan\beta=\frac{1}{2m+1}$ then $\alpha+\beta$ is equal to

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Phrases Previous Year PYQ Phrases NIMCET 2013 PYQ

Solution

Qus : 32 Phrases PYQ

If $\sin x=\sin y$ and $\cos x=\cos y$, then the value of x-y is

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Phrases Previous Year PYQ Phrases NIMCET 2024 PYQ

Solution

Given:

\[ \sin x = \sin y \quad \text{and} \quad \cos x = \cos y \]

✳ Step 1: Use the identity for sine

\[ \sin x = \sin y \Rightarrow x = y + 2n\pi \quad \text{or} \quad x = \pi - y + 2n\pi \]

✳ Step 2: Use the identity for cosine

\[ \cos x = \cos y \Rightarrow x = y + 2m\pi \quad \text{or} \quad x = -y + 2m\pi \]

? Combine both conditions

For both $ \sin x = \sin y $ and $ \cos x = \cos y $ to be true, the only consistent solution is:

\[ x = y + 2n\pi \Rightarrow x - y = 2n\pi \]

✅ Final Answer:

\[ \boxed{x - y = 2n\pi \quad \text{for } n \in \mathbb{Z}} \]

Qus : 33 Phrases PYQ

If $a_1, a_2, a_3,...a_n$, are in Arithmetic Progression with common difference d, then the sum $(sind) (cosec a_1 . cosec a_2+cosec a_2.cosec a_2+...+cosec a_{n-1}.cosec a_n)$ is equal to

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Phrases Previous Year PYQ Phrases NIMCET 2022 PYQ

Solution

Qus : 34 Phrases PYQ

In a ΔABC, if $\tan ^2\frac{A}{2}+\tan ^2\frac{B}{2}+\tan ^2\frac{C}{2}=k$ , then k is always

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Phrases Previous Year PYQ Phrases NIMCET 2021 PYQ

Solution

Qus : 35 Phrases PYQ

The general value of $\theta$, satisfying the equation $\sin \theta=\frac{-1}{2},\, \tan \theta=\frac{1}{\sqrt[]{3}}$

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Phrases Previous Year PYQ Phrases NIMCET 2021 PYQ

Solution

Qus : 36 Phrases PYQ

If $\cos\alpha+\cos\beta=a$, $\sin\alpha+\sin\beta=b$ and $\theta$ is the arithmetic mean between $\alpha$ and $\beta$, then $\sin2\theta+\cos2\theta$ is equal to

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Phrases Previous Year PYQ Phrases NIMCET 2008 PYQ

Solution

$a^2+b^2=(\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2$ $=2+2\cos(\alpha-\beta)=4\cos^2\frac{\alpha-\beta}{2}$ $\Rightarrow \cos(\alpha-\beta)=\dfrac{a^2+b^2}{2}-1$ Since $\theta=\dfrac{\alpha+\beta}{2}$, $\sin2\theta+\cos2\theta=\dfrac{a^2-b^2}{a^2+b^2}$ Answer: $\boxed{\dfrac{a^2-b^2}{a^2+b^2}}$

Qus : 37 Phrases PYQ

If $\dfrac{\tan x}{2} = \dfrac{\tan y}{3} = \dfrac{\tan z}{5}$ and $x + y + z = \pi$, then the value of $\tan^2 x + \tan^2 y + \tan^2 z$ is

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Phrases Previous Year PYQ Phrases NIMCET 2018 PYQ

Solution

Qus : 38 Phrases PYQ

If $(1+\tan1^\circ)(1+\tan2^\circ)\cdots(1+\tan45^\circ)=2^n$, then the value of $n$ is

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Phrases Previous Year PYQ Phrases NIMCET 2008 PYQ

Solution

Using identity $(1+\tan\theta)(1+\tan(45^\circ-\theta))=2$ There are $22$ such pairs from $1^\circ$ to $44^\circ$ and $(1+\tan45^\circ)=2$ So $2^{22}\times2=2^{23}$ Hence $n=23$ Answer: $\boxed{23}$

Qus : 39 Phrases PYQ

The value of $\sin12^\circ\sin48^\circ\sin54^\circ$ is

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Phrases Previous Year PYQ Phrases NIMCET 2008 PYQ

Solution

Using standard trigonometric product identities, $\sin12^\circ\sin48^\circ\sin54^\circ=\sin^330^\circ$ Answer: $\boxed{\sin^330^\circ}$

Qus : 40 Phrases PYQ

If tan x = - 3/4 and 3π/2 < x < 2π, then the value of sin2x is

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Phrases Previous Year PYQ Phrases NIMCET 2017 PYQ

Solution

Qus : 41 Phrases PYQ

The value of $\tan 9{^{\circ}}-\tan 27{^{\circ}}-\tan 63{^{\circ}}+\tan 81{^{\circ}}$ is equal to

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Phrases Previous Year PYQ Phrases NIMCET 2021 PYQ

Solution

Qus : 42 Phrases PYQ

If cosθ = 4/5 and cosϕ = 12/13, θ and ϕ both in the fourth quadrant, the value of cos( θ + ϕ )is

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Phrases Previous Year PYQ Phrases NIMCET 2017 PYQ

Solution

Qus : 43 Phrases PYQ

The value of sin36^o is

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Phrases Previous Year PYQ Phrases NIMCET 2017 PYQ

Solution

Qus : 44 Phrases PYQ

Express (cos 5x – cos7x) as a product of sines or cosines or sines and cosines,

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Phrases Previous Year PYQ Phrases NIMCET 2017 PYQ

Solution

Qus : 45 Phrases PYQ

If $32\, \tan ^8\theta=2\cos ^2\alpha-3\cos \alpha$ and $3\, \cos \, 2\theta=1$, then the general value of $\alpha$ =

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Phrases Previous Year PYQ Phrases NIMCET 2021 PYQ

Solution

Qus : 46 Phrases PYQ

If |k|=5 and 0° ≤ θ ≤ 360°, then the number of distinct solutions of 3cos⁡θ + 4sin⁡θ = k is

NIMCET 2021

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Phrases Previous Year PYQ Phrases NIMCET 2021 PYQ

Solution

Qus : 47 Phrases PYQ

If $A-B=\frac{\pi}{4}$, then (1 + tan A)(1 – tan B) is equal to

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Phrases Previous Year PYQ Phrases NIMCET 2012 PYQ

Solution

Qus : 48 Phrases PYQ

If $B=sin^2 y+cos^4 y$, then for all real y

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Phrases Previous Year PYQ Phrases NIMCET 2025 PYQ

Solution

Let $B = \sin^{2}y + \cos^{4}y$.

Put $t = \cos^{2}y$, so $0 \le t \le 1$.

Then $\sin^{2}y = 1 - t$.

So, $B = (1 - t) + t^{2} = t^{2} - t + 1$.

Minimum of the quadratic $t^{2} - t + 1$ occurs at $t = \dfrac{1}{2}$:

$B_{\min} = \dfrac{1}{4} - \dfrac{1}{2} + 1 = \dfrac{3}{4}$.

Maximum occurs at $t = 0$ or $t = 1$:

$B_{\max} = 1$.

Hence, $B \in \left[\dfrac{3}{4},, 1\right]$.

Qus : 49 Phrases PYQ

Value of $\sqrt{3}\cos 20^\circ - 4\cos 20^\circ$ is:

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Phrases Previous Year PYQ Phrases NIMCET 2010 PYQ

Solution

$\cos 20^\circ(\sqrt{3}-4)$ is negative because $(\sqrt{3}-4)<0$. Compute approximate: $\cos20^\circ \approx 0.94$ $\sqrt{3}-4 \approx -2.268$ Product ≈ $-2.13$ Which is none of the options 1, -1, 0.

Qus : 50 Phrases PYQ

The maximum value of $\sin x+\sin(x+1)$ is $k \cos \frac{1}{2}$. Then the value of k is

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Phrases Previous Year PYQ Phrases NIMCET 2025 PYQ

Solution

We use the identity

$\sin A + \sin B = 2 \sin \dfrac{A+B}{2} \cos \dfrac{A-B}{2}$.

So, $\sin x + \sin(x+1) = 2 \sin\left(x+\dfrac12\right)\cos\dfrac12$.

Maximum value of $\sin(x+\tfrac12)$ is $1$.

Therefore maximum of the expression is: $2 \cos\dfrac12$.

Given maximum $= k \cos\dfrac12$, so $k = 2$.

Qus : 51 Phrases PYQ

If $\sin\theta = 3\sin(\theta + 2\alpha)$, then value of $\tan(\theta + \alpha) + 2\tan\alpha$ is

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Phrases Previous Year PYQ Phrases NIMCET 2018 PYQ

Solution

Qus : 52 Phrases PYQ

What is the general solution of the equation $\tan \theta + \cot \theta = 2$ ?

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Phrases Previous Year PYQ Phrases NIMCET 2025 PYQ

Solution

Given equation: $ \tan\theta + \cot\theta = 2 $.

Rewrite $\cot\theta$:

$ \tan\theta + \dfrac{1}{\tan\theta} = 2 $.

Let $t = \tan\theta$.

Then: $ t + \dfrac{1}{t} = 2 $.

Multiply by $t$:

$ t^2 + 1 = 2t $.

Rearrange:

$ t^2 - 2t + 1 = 0 $.

$ (t - 1)^2 = 0 $.

So: $ t = 1 $.

Thus: $ \tan\theta = 1 $.

General solution: $ \theta = \dfrac{\pi}{4} + n\pi,\; n \in \mathbb{Z}. $

Qus : 53 Phrases PYQ

If $a\, \cos \theta+b\, \sin \, \theta=2$ and $a\, \sin \, \theta-b\, \cos \, \theta=3$ , then ${a}^{2^{}}+{b}^2=$

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Phrases Previous Year PYQ Phrases NIMCET 2021 PYQ

Solution

Qus : 54 Phrases PYQ

If $\cos^2(10°)\cos(20°)\cos(40°)\cos(50°) \cos(70°) = \alpha+\frac{\sqrt{3}}{16} \cos(10°)$, then $3\alpha^{-1}$ is equal to

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Phrases Previous Year PYQ Phrases NIMCET 2025 PYQ

Solution

Qus : 55 Phrases PYQ

The value of tan 1° tan 2° tan 3° ... tan 89° is:

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Phrases Previous Year PYQ Phrases NIMCET 2014 PYQ

Solution

Qus : 56 Phrases PYQ

If $P=sin^{20} \theta + cos^{48} \theta $ then the inequality that holds for all values of is

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Phrases Previous Year PYQ Phrases NIMCET 2015 PYQ

Solution

Qus : 57 Phrases PYQ

If $sin x + a cos x = b$, then $|a sin x - cos x|$ is:

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Phrases Previous Year PYQ Phrases NIMCET 2014 PYQ

Solution

Qus : 58 Phrases PYQ

If $\cos\theta = \dfrac{5}{13},; \dfrac{3\pi}{2} < \theta < 2\pi$, then $\tan 2\theta$ is

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Phrases Previous Year PYQ Phrases NIMCET 2016 PYQ

Solution

Qus : 59 Phrases PYQ

If $0 < x < \pi $ and $cos x + sin x = \frac{1}{2}$ , then the value of tan x is

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Phrases Previous Year PYQ Phrases NIMCET 2015 PYQ

Solution

Qus : 60 Phrases PYQ

If tan A - tan B = x and cot B - cot A = y, then cot (A - B) is equal to

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Phrases Previous Year PYQ Phrases NIMCET 2014 PYQ

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Qus : 61 Phrases PYQ

The value of sin 20° sin 40° sin 80° is

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Phrases Previous Year PYQ Phrases NIMCET 2014 PYQ

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Qus : 62 Phrases PYQ

In a right angled triangle, the hypotenuse is four times the perpendicular drawn to it from the opposite vertex. The value of one of the acute angles is

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Phrases Previous Year PYQ Phrases NIMCET 2015 PYQ

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Qus : 63 Phrases PYQ

If $\prod ^n_{i=1}\tan ({{\alpha}}_i)=1\, \forall{{\alpha}}_i\, \in\Bigg{[}0,\, \frac{\pi}{2}\Bigg{]}$ where i=1,2,3,...,n. Then maximum value of $\prod ^n_{i=1}\sin ({{\alpha}}_i)$.

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Phrases Previous Year PYQ Phrases NIMCET 2023 PYQ

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Qus : 64 Phrases PYQ

Solve the equation sin²x - sinx - 2 = 0 for for x on the interval 0 ≤ x < 2π

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Phrases Previous Year PYQ Phrases NIMCET 2020 PYQ

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Qus : 65 Phrases PYQ

If $\frac{tanx}{2}=\frac{tanx}{3}=\frac{tanx}{5}$ and x + y + z = π, then the value of tan²x + tan²y + tan²z is

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Phrases Previous Year PYQ Phrases NIMCET 2020 PYQ

Solution

Qus : 66 Phrases PYQ

Find the value of sin 12°sin 48°sin 54°

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Phrases Previous Year PYQ Phrases NIMCET 2020 PYQ

Solution

Qus : 67 Phrases PYQ

If cos x = tan y , cot y = tan z and cot z = tan x, then sinx =

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Phrases Previous Year PYQ Phrases NIMCET 2020 PYQ

Solution

Qus : 68 Phrases PYQ

The value of $tan(\frac{7\pi}{8})$ is

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Phrases Previous Year PYQ Phrases NIMCET 2015 PYQ

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Qus : 69 Phrases PYQ

The value of $\cos 20^\circ + \cos 100^\circ + \cos 140^\circ$ is

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Phrases Previous Year PYQ Phrases NIMCET 2016 PYQ

Solution

We need to find: $\cos 20^\circ + \cos 100^\circ + \cos 140^\circ$
Using sum-to-product formula on first two terms:
$\cos 20^\circ + \cos 100^\circ = 2\cos\left(\dfrac{20^\circ + 100^\circ}{2}\right)\cos\left(\dfrac{100^\circ - 20^\circ}{2}\right)$
$= 2\cos 60^\circ \cos 40^\circ$
$= 2 \times \dfrac{1}{2} \times \cos 40^\circ$
$= \cos 40^\circ$
So the expression becomes:
$\cos 40^\circ + \cos 140^\circ$
Again applying sum-to-product formula:
$= 2\cos\left(\dfrac{40^\circ + 140^\circ}{2}\right)\cos\left(\dfrac{140^\circ - 40^\circ}{2}\right)$
$= 2\cos 90^\circ \cos 50^\circ$
$= 2 \times 0 \times \cos 50^\circ$
$= 0$
$\therefore \boxed{\cos 20^\circ + \cos 100^\circ + \cos 140^\circ = 0}$

Qus : 70 Phrases PYQ

The value of

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Phrases Previous Year PYQ Phrases NIMCET 2020 PYQ

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Qus : 71 Phrases PYQ

The value of sin 10°sin 50°sin 70° is

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Phrases Previous Year PYQ Phrases NIMCET 2020 PYQ

Solution

sin10° sin50° sin70°

= sin10° sin(60°−10°) sin(60°+10°)

= 1/4 sin3x10°

=1/4x1/2=1/8

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