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MCA NIMCET Previous Year Questions (PYQs)

MCA NIMCET Trigonometry PYQ

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





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Solution















MCA NIMCET 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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Solution

According to the given information, the figure should be as follows.  
Let the height of tower = h


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





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Solution

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





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Solution

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





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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:   \( \boxed{2 \text{ points}} \)

MCA NIMCET 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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Solution

The overlapping area of a square rotated by $45^\circ$ is: Common area $= 2(\sqrt{2}-1)a^2$
MCA NIMCET PYQ
The equation $\sin^4 x + \cos^4 x + \sin 2x + \alpha = 0$ is solvable for:





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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: 
${-\dfrac{3}{2} \le \alpha \le \dfrac{1}{2}}$
MCA NIMCET PYQ
If $\sin^2 x = 1 - \sin x$, then $\cos^4 x + \cos^2 x$ is equal to:





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





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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: ${\dfrac{3}{4} \le A \le 1}$
MCA NIMCET PYQ
The value of $ \sin 30^\circ \cos 45^\circ + \cos 30^\circ \sin 45^\circ $





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Solution

Expression is $\sin(30^\circ + 45^\circ)$ $ = \sin 75^\circ = \frac{\sqrt{6}+\sqrt{2}}{4} = \frac{\sqrt{3}+1}{2\sqrt{2}} $
MCA NIMCET 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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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 = $a$.
MCA NIMCET PYQ
The general solution of $ \sqrt{3}\cos x + \sin x = 3 $ is:





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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.
MCA NIMCET 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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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 \[ \min(\tan A+\tan B)=\frac{\frac12}{\frac{2+\sqrt3}{4}} =\frac{2}{2+\sqrt3} =\boxed{\,4-2\sqrt3\,}. \]

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





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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} $
MCA NIMCET PYQ
If $ \cos(\alpha+\beta)=\frac{4}{5}$ and  $\sin(\alpha-\beta)=\frac{5}{13},\quad 0<\alpha,\beta<\frac{\pi}{4},$ then $ \tan(2\alpha)$ is equal to





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Solution

Given: 
$\cos(\alpha+\beta)=\frac{4}{5}$ 
Since $0<\alpha,\beta<\frac{\pi}{4}$, 
therefore $\alpha+\beta$ is an acute angle. 

$\sin(\alpha+\beta)=\sqrt{1-\cos^2(\alpha+\beta)}$ 
$=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$ 

Therefore, $\tan(\alpha+\beta)=\frac{3}{4}$ 

Also, $\sin(\alpha-\beta)=\frac{5}{13}$ 

Since the sine value is positive, $\alpha-\beta>0$. 

Therefore, $\cos(\alpha-\beta)=\sqrt{1-\sin^2(\alpha-\beta)}$ 

$=\sqrt{1-\frac{25}{169}}=\frac{12}{13}$ 

Hence, $\tan(\alpha-\beta)=\frac{5}{12}$ 

Now, $2\alpha=(\alpha+\beta)+(\alpha-\beta)$ 

Using the tangent addition formula: 

$\tan(2\alpha)=\frac{\tan(\alpha+\beta)+\tan(\alpha-\beta)}{1-\tan(\alpha+\beta)\tan(\alpha-\beta)}$ 

$=\frac{\frac{3}{4}+\frac{5}{12}}{1-\left(\frac{3}{4}\right)\left(\frac{5}{12}\right)}$ 

$=\frac{\frac{14}{12}}{1-\frac{15}{48}}$ 

$=\frac{\frac{7}{6}}{\frac{33}{48}}$ 

$=\frac{7}{6}\times\frac{48}{33}$ 

$=\frac{56}{33}$ 

Therefore, the correct answer is: $\frac{56}{33}$
MCA NIMCET PYQ
If $A-B=\frac{\pi}{4}$, then $(1+\tan A)(1-\tan B)$ is equal to





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Solution

We know, $\tan(A-B)=\tan\frac{\pi}{4}=1$ 

Using the formula, 
$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$ 
Therefore, $\frac{\tan A-\tan B}{1+\tan A\tan B}=1$ 
$\tan A-\tan B=1+\tan A\tan B$ 
Now, $(1+\tan A)(1-\tan B)$ 
$=1+\tan A-\tan B-\tan A\tan B$ 
$=1+(\tan A-\tan B)-\tan A\tan B$ 
$=1+(1+\tan A\tan B)-\tan A\tan B$ 
$=2$ 
Therefore, the correct answer is: 2
MCA NIMCET PYQ
Which of the following is correct?





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Solution

Here, $\sin 1^\circ$ means the angle is measured in degrees, while $\sin 1$ means the angle is measured in radians. Convert $1$ radian into degrees: $1\text{ radian}=\frac{180^\circ}{\pi}\approx57.3^\circ$ 

Therefore, $1^\circ<1\text{ radian}<90^\circ$ 
Since the sine function is increasing in the first quadrant, $\sin 1^\circ<\sin 1$ 

Therefore, the correct answer is: $\sin 1^\circ<\sin 1$
MCA NIMCET PYQ
If $\sin(\pi\cos\theta)=\cos(\pi\sin\theta)$, then $\sin 2\theta$ is equal to





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Solution

Given, $\sin(\pi\cos\theta)=\cos(\pi\sin\theta)$ 

Using, $\cos x=\sin\left(\frac{\pi}{2}-x\right)$ 

Therefore, $\sin(\pi\cos\theta)=\sin\left(\frac{\pi}{2}-\pi\sin\theta\right)$ 

This gives the following two possible cases. 

Case 1: $\pi\cos\theta=\frac{\pi}{2}-\pi\sin\theta$ 
$\cos\theta+\sin\theta=\frac{1}{2}$ 
Squaring both sides, $(\cos\theta+\sin\theta)^2=\frac{1}{4}$ 

$\cos^2\theta+\sin^2\theta+2\sin\theta\cos\theta=\frac{1}{4}$ 

$1+\sin 2\theta=\frac{1}{4}$ 
$\sin 2\theta=-\frac{3}{4}$ 

Case 2: 
$\pi\cos\theta=\pi-\left(\frac{\pi}{2}-\pi\sin\theta\right)$ 
$\cos\theta-\sin\theta=\frac{1}{2}$ 
Squaring both sides, 
$(\cos\theta-\sin\theta)^2=\frac{1}{4}$ 
$1-\sin 2\theta=\frac{1}{4}$ 
$\sin 2\theta=\frac{3}{4}$ 
Hence, $\sin 2\theta=\pm\frac{3}{4}$ 
 Therefore, the correct answer is: $\boxed{\text{(a) }\pm\frac{3}{4}}$
MCA NIMCET 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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Solution

MCA NIMCET PYQ
The equation $(\cos p-1)x^2+(\cos p)x+\sin p=0$ where $x$ is a variable, has real roots. Then, the interval of $p$ is





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Solution

For the quadratic equation to have real roots, its discriminant must be non-negative. 

Here, $a=\cos p-1,\qquad b=\cos p,\qquad c=\sin p$ 

Therefore, $\Delta=b^2-4ac$ 

$\Delta=\cos^2p-4(\cos p-1)\sin p$ 

$\Delta=\cos^2p+4(1-\cos p)\sin p$ 

For 0<p<0$\pi$, $\sin p >0$ and $1-\cos p>0$ 

Therefore, $\cos^2p+4(1-\cos p)\sin p>0$ 

Hence, the equation has two real roots for $p\in(0,\pi)$.
MCA NIMCET PYQ
If $\sin x,\ \cos x,\ \tan x$ are in GP, find $\cot 6x - \cot 2x$.





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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 $1$.
MCA NIMCET PYQ
Solve: $ 2\sin^{2}\theta - 3\sin\theta - 2 = 0$





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





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Solution

MCA NIMCET 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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Solution

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





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Solution

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





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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 $$\csc\theta=\frac{5}{4}.$$
MCA NIMCET PYQ
The solution of the equation ${4\cos }^2x+6{\sin }^2x=5$ are





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Solution

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





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Solution

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





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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:

\[ \boxed{-1} \]

MCA NIMCET 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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Solution

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





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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:

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

MCA NIMCET 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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Solution

MCA NIMCET 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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Solution

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





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Solution

MCA NIMCET 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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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: $\boxed{\dfrac{a^2-b^2}{a^2+b^2}}$
MCA NIMCET 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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Solution

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





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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: $\boxed{23}$
MCA NIMCET PYQ
The value of $\sin12^\circ\sin48^\circ\sin54^\circ$ is





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Solution

Using standard trigonometric product identities, $\sin12^\circ\sin48^\circ\sin54^\circ=\sin^330^\circ$ Answer: $\boxed{\sin^330^\circ}$
MCA NIMCET PYQ
If tan x = - 3/4 and 3π/2 < x < 2π, then the value of sin2x is





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Solution


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





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Solution

MCA NIMCET 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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Solution

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





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Solution

MCA NIMCET 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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Solution

MCA NIMCET 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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Solution

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





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Solution

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





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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, $B \in \left[\dfrac{3}{4},, 1\right]$.
MCA NIMCET PYQ
Value of $\sqrt{3}\cos 20^\circ - 4\cos 20^\circ$ is:





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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.
MCA NIMCET 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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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 $k = 2$. 
MCA NIMCET PYQ
If $\sin\theta = 3\sin(\theta + 2\alpha)$, then value of $\tan(\theta + \alpha) + 2\tan\alpha$ is





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





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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}. $
MCA NIMCET 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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MCA NIMCET 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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MCA NIMCET PYQ
The value of tan 1° tan 2° tan 3° ... tan 89° is:





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Solution

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





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





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





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





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





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MCA NIMCET PYQ
The value of sin 20° sin 40° sin 80° is





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MCA NIMCET 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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MCA NIMCET 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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MCA NIMCET PYQ
Solve the equation sin2 x - sinx - 2 = 0 for for x on the interval 0 ≤ x < 2π





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MCA NIMCET PYQ
Given that $\cos 6x=a\cos^6x+b\cos^4x+c\cos^2x+d$ for any real number $x$, find the value of $a+b+c$.





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Solution

Given: $\cos 6x=a\cos^6x+b\cos^4x+c\cos^2x+d$ Put $x=0$ 
$\cos 0=a\cos^6 0+b\cos^4 0+c\cos^2 0+d$ 
$1=a+b+c+d$ ...(1) 

Put $x=\frac{\pi}{2}$ 

$\cos 3\pi=a\cos^6\frac{\pi}{2}+b\cos^4\frac{\pi}{2}+c\cos^2\frac{\pi}{2}+d$ 
$-1=d$ ...(2) 

From (1), 
$a+b+c+d=1$ 
$a+b+c-1=1$ 
$a+b+c=2$ 
Answer: $2$
MCA NIMCET 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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MCA NIMCET PYQ
Find the value of sin 12°sin 48°sin 54°





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





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





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Solution

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





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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:
$= 2\cos\left(\dfrac{40^\circ + 140^\circ}{2}\right)\cos\left(\dfrac{140^\circ - 40^\circ}{2}\right)$
$= 2\cos 90^\circ \cos 50^\circ$
$= 2 \times 0 \times \cos 50^\circ$
$= 0$
$\therefore \boxed{\cos 20^\circ + \cos 100^\circ + \cos 140^\circ = 0}$
MCA NIMCET PYQ
The value of  is





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Solution

= 

= 

= 

= 

= 

= 

= 

= 
MCA NIMCET PYQ
The value of sin 10°sin 50°sin 70° is





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Solution

sin10° sin50° sin70°
= sin10° sin(60°−10°) sin(60°+10°)
= 1/4 sin3x10°
=1/4x1/2=1/8

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