Rd Sharma 2022 _mcqs for Class 10 Maths Chapter 11 - Trigonometric Identities

Answer:

Given: x = 2 sin² θ and y = 2 cos² θ + 1

$\begin{array}{rcl}x+y& =& 2{\sin }^2\theta +2{\cos }^2\theta +1\\ & =& 2\left({\sin }^2\theta +{\cos }^2\theta \right)+1\\ & =& 2\times 1+1 \left[\because {\sin }^2\theta +{\cos }^2\theta =1\right]\\ & =& 3\end{array}$

Hence, the correct answer is option (a).

Answer:

Given: tanα + cotα = 2,

Now, 
$$\begin{gathered} \tan \alpha +\cot \alpha =2 \\ \Rightarrow \tan \alpha +\frac{1}{\tan \alpha }=2 \\ \Rightarrow \frac{1+{\tan }^2\alpha }{\tan \alpha }=2 \\ \Rightarrow \frac{{\sec }^2\alpha }{\tan \alpha }=2 \left(\because 1+{\tan }^2\theta ={\sec }^2\theta \right) \\ \Rightarrow \frac{1}{{\cos }^2\alpha }\times \frac{\cos \alpha }{\sin \alpha }=2 \left(\because \sec \theta =\frac{1}{\cos \theta }\right) \\ \Rightarrow \frac{1}{2\sin \alpha \cos \alpha }=1 \\ \Rightarrow \sin 2\alpha =1 \left(\because 2\sin \theta \cos \theta =\sin 2\theta \right) \\ \Rightarrow \sin 2\alpha =\sin 90° \left(\because \sin 90°=1\right) \\ \Rightarrow 2\alpha =90° \\ \Rightarrow \alpha =45° \end{gathered}$$

$\begin{array}{rcl}\therefore {\tan }^{2020}\alpha +{\cot }^{2020}\alpha & =& {\tan }^{2020}45°+{\cot }^{2020}45°\\ & =& {\left(1\right)}^{2020}+{\left(1\right)}^{2020} \left(\because \tan 45°=1 \mathrm{and} \cot 45°=1\right) \\ & =& 2\end{array}$

Hence, the correct answer is option (b).

Answer:

The given expression is .

Multiplying both the numerator and denominator under the root by , we have

Therefore, the correct option is (a).

Answer:

The given expression is .

Multiplying both the numerator and denominator under the root by, we have

Therefore, the correct choice is (b).

Answer:

The given expression is .

Multiplying both the numerator and denominator under the root by , we have

Therefore, the correct option is (c).

Answer:

The given expression is .

Simplifying the given expression, we have

Therefore, the correct option is (c).

Answer:

The given expression is .

Simplifying the given expression, we have

Therefore, the correct option is (c).

Answer:

Given:

So,

We know that,

Therefore,

Hence, the correct option is (a).

Answer:

Given:

So,

We know that,

Therefore,

Hence, the correct option is (d).

Answer:

The given expression is .

Hence, the correct option is (a).

Answer:

Given:

Now,

Hence, the correct option is (d).

Answer:

Given:

We know that,

Therefore,

Hence, the correct option is (b).

Answer:

The given expression is .

Simplifying the given expression, we have

Therefore, the correct option is (d).

Answer:

Given:

 

Therefore, the correct option is (d).

Answer:

We know that, $\sin \left(90-\theta \right)=\cos \theta$.
So, $\sin \left(45°+\theta \right)=\cos \left[90-\left(45°+\theta \right)\right]=\cos \left(45°-\theta \right)$
$$\begin{gathered} \therefore \sin \left(45°+\mathrm{\theta }\right)-\cos \left(45°-\mathrm{\theta }\right) \\ =\cos \left(45°-\mathrm{\theta }\right)-\cos \left(45°-\mathrm{\theta }\right) \\ =0 \end{gathered}$$
Hence, the correct answer is option (b).

Answer:

In a right angled triangle ABC, $\angle \mathrm{C}$ is a right angle.
We know that, the sum of angles of a triangle is 180º.
$$\begin{gathered} \therefore \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180° \\ \Rightarrow \angle \mathrm{A}+\angle \mathrm{B}+90°=180° \\ \Rightarrow \angle \mathrm{A}+\angle \mathrm{B}=90° \end{gathered}$$
$\therefore \cos \left(\mathrm{A}+\mathrm{B}\right)=\cos 90°=0$
Hence, the correct answer is option (a).

Answer:

Given:

We know that,

Now,

Subtracting the second equation from the first equation, we get

Therefore, the correct choice is (d).

Answer:

Given:

We know that,

Now,

Adding the two equations, we get

Therefore, the correct choice is (b).

Answer:

The given expression is .

Taking common from both the terms, we have

Disclaimer: The options given in (c) and (d) are same by the commutative property of addition.

Therefore, the correct options are (c) or (d).

Answer:

The given expression is .

Factorising the given expression, we have

Therefore, the correct option is (b).

Answer:

The given expression is

Simplifying the given expression, we have

Therefore, the correct option is (b).

Answer:

The given expression is

Simplifying the given expression, we have

Therefore, the correct option is (b).

Answer:

Given: sin (A – B) = 0 and 2 cos (A + B) – 1 = 0 where A and B are acute angles.

Since sin (A – B) = 0
⇒ sin (A – B) = sin 0º          (∵ sin 0º = 0)
⇒ A – B = 0                      ..... (1)

2 cos (A + B) – 1 = 0
⇒ cos (A + B) = $\frac{1}{2}$
⇒ cos (A + B) = cos 60^∘       (∵ cos 60^∘ = $\frac{1}{2}$)     
⇒ A + B = 60^∘                 .....(2)

Adding (1) and (2), we get
B = 30^∘ and A = 30^∘

Hence, the correct answer is option (b).

Answer:

Given: sin θ – cos θ = 0
$$\begin{gathered} \Rightarrow \sin \theta =\cos \theta \\ \Rightarrow \frac{\sin \theta }{\cos \theta }=1 \\ \Rightarrow \tan \theta =1 \\ \Rightarrow \tan \theta =\tan 45° \left(\because \tan 45°=1\right) \\ \Rightarrow \theta =45° \end{gathered}$$

$\begin{array}{rcl}\therefore {\sin }^4\theta +{\cos }^4\theta & =& {\left(\sin 45°\right)}^4+{\left(\cos 45°\right)}^4\\ & =& {\left(\frac{1}{\sqrt{2}}\right)}^4+{\left(\frac{1}{\sqrt{2}}\right)}^4 \left(\because \sin 45°=\cos 45°=\frac{1}{\sqrt{2}} \right)\\ & =& \frac{1}{4}+\frac{1}{4}\\ & =& \frac{1}{2}\end{array}$

Hence, the correct answer is option (c).

Answer:

It is given that,
$$\begin{gathered} \cos 9\theta =\sin \theta , 9\theta <90° \\ \Rightarrow \sin \left(90°-9\theta \right)=\sin \theta \left[\sin \left(90°-\theta \right)=\cos \theta \right] \\ \Rightarrow 90°-9\theta =\theta \\ \Rightarrow 10\theta =90° \\ \Rightarrow \theta =9° \\ \mathrm{Therefore}, \tan 6\theta =\tan 54°. \end{gathered}$$
Disclaimer: Answer of the given question is not matching with the options provided in the textbook.

Answer:

It is given that,
$$\begin{gathered} \cos \left(\alpha +\beta \right)=0 \\ \Rightarrow \cos \left(\alpha +\beta \right)=\cos 90° \left(\cos 90°=0\right) \\ \Rightarrow \alpha +\beta =90° \\ \Rightarrow \alpha =90°-\beta \\ \mathrm{Now}, \mathrm{put} \alpha =90°-\beta \mathrm{in} \sin \left(\alpha -\beta \right). \\ \therefore \sin \left(\alpha -\beta \right) \\ =\sin \left(90°-\beta -\beta \right) \\ =\sin \left(90°-2\beta \right) \\ =\cos 2\beta \left[\sin \left(90°-\theta \right)=\cos \theta \right] \end{gathered}$$
Hence, the correct answer is option (b).

Answer:

Given: 1 + sin²α = 3 sinα cosα         .....(1)

Dividing both sides of (1) by sin²α,
$$\begin{gathered} \Rightarrow \frac{1+{\sin }^2\alpha }{{\sin }^2\alpha }=\frac{3\sin \alpha \cos \alpha }{{\sin }^2\alpha } \\ \Rightarrow {\mathrm{cosec}}^2\alpha +1=3\cot \alpha \\ \Rightarrow 1+{\cot }^2\alpha +1=3\cot \alpha \\ \Rightarrow {\cot }^2\alpha -3\cot \alpha +2=0 \\ \Rightarrow \left(\cot \alpha -1\right)\left(\cot \alpha -2\right)=0 \\ \Rightarrow \cot \alpha =1 \mathrm{or} 2 \end{gathered}$$

Hence, the correct answer is option (c).

Answer:

The given expression is .

Simplifying the given expression, we have

Therefore, the correct option is (b).

Answer:

The given expression is .

Simplifying the given expression, we have

Therefore, the correct option is (c).

Answer:

Given:

Squaring and then adding the above two equations, we have

Hence, the correct option is (c).

Answer:

Given:

Squaring both the equations and then subtracting the second from the first, we have

 
 

Hence, the correct option is (b).

Answer:

Given:

Squaring and adding these equations, we get

Hence, the correct option is (a).

Answer:

Given:

Now,

Hence, the correct option is (b).

Answer:

Given:

Squaring and adding these equations, we have

Hence, the correct option is (d).

Answer:

Given:

So,

 
 

Hence, the correct option is (c).

Answer:

Given:

Squaring on both sides, we have

 



Hence, the correct option is (b).

Answer:

Given: 2sin²β – cos²β = 2

$$\begin{gathered} \Rightarrow 2{\sin }^2\beta -\left(1-{\sin }^2\beta \right)=2 \left(\because 1-{\sin }^2\theta ={\cos }^2\theta \right) \\ \Rightarrow 3{\sin }^2\beta =3 \\ \Rightarrow {\sin }^2\beta =1 \\ \Rightarrow \sin \beta =1 \\ \Rightarrow \sin \beta =\sin 90° \left(\because \sin 90°=1\right) \\ \Rightarrow \beta =90° \end{gathered}$$

Hence, the correct answer is option (b).

Answer:

Statement-2 (Reason): For 0 < θ ≤ 90°, tan (90° – θ) = cotθ and tan45° = 1.
For 0 < θ ≤ 90°, tan (90° – θ) = cotθ and tan45° = 1.
Thus, statement-2 is true.

Statement-1 (Assertion): The value of the product P = tan1° tan2° tan3°,......,tan89° is 1.
P = tan1° tan2° tan3°,......,tan89°
= tan1° tan2° tan3°...tan45°tan46°...tan88°tan89°      
= tan1° tan2° tan3°...1.cot44°...cot2°cot1°                   (From Statement-1)
= 1

Thus, statement-1 is true.

So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

Answer:

Statement-2 (Reason): The value of cos 90° is zero.
Since, cos 90° = 0.
Thus, Statement-2 is true.

Statement-1 (Assertion): The value of the product of P = cos1° cos2°..... cos179° is zero.
P = cos1° cos2°..... cos179°
= cos1° cos2°...cos90°... cos179°
= cos1° cos2°...0... cos179°              (From Statement-1)
= 0

Thus, statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
Hence, the correct answer is option (a).

Answer:

Statement-2 (Reason): cot²θ – cosec²θ = 1.
Using trigonometric identity, 1+ cot²θ = cosec²θ.
⇒ cot²θ – cosec²θ = –1
Thus, Statement-2 is false.

Statement-1 (Assertion): For 0 < θ ≤ 90°, cosecθ – cotθ and cosecθ + cotθ are reciprocal of each other.
$\begin{array}{rcl}\mathrm{cosec\theta }+\mathrm{cot\theta }& =& \left(\mathrm{cosec\theta }+\mathrm{cot\theta }\right)\times \frac{\mathrm{cosec\theta }-\mathrm{cot\theta }}{\mathrm{cosec\theta }-\mathrm{cot\theta }}\\ & =& \frac{\left(\mathrm{cosec\theta }+\mathrm{cot\theta }\right)\left(\mathrm{cosec\theta }-\mathrm{cot\theta }\right)}{\mathrm{cosec\theta }-\mathrm{cot\theta }}\\ & =& \frac{{\mathrm{cosec}}^2\mathrm{\theta }-{\cot }^2\mathrm{\theta }}{\mathrm{cosec\theta }-\mathrm{cot\theta }}\\ & =& \frac{1}{\mathrm{cosec\theta }-\mathrm{cot\theta }} \left(\because 1+{\cot }^2\mathrm{\theta }={\mathrm{cosec}}^2\mathrm{\theta }\right)\end{array}$

For 0 < θ ≤ 90°, cosecθ – cotθ and cosecθ + cotθ are reciprocal of each other.
Thus, Statement-1 is true.

So, Statement-1 is true, Statement-2 is false.

Hence, the correct answer is option (c).

Answer:

Statement-2 (Reason): cosec²θ – cot²θ = 1.
Using trigonometric identity, 1+ cot²θ = cosec²θ.
⇒ 1 = cosec²θ – cot²θ
Thus, Statement-2 is true.

Statement-1 (Assertion): For 0 ≤ θ < 90°, secθ + tanθ and secθ – tanθ are reciprocal of each other.
$\begin{array}{rcl}\sec \theta +\tan \theta & =& \left(\sec \theta +\tan \theta \right)\times \frac{\sec \theta -\tan \theta }{\sec \theta -\tan \theta }\\ & =& \frac{\left(\sec \theta +\tan \theta \right)\left(\sec \theta -\tan \theta \right)}{\sec \theta -\tan \theta }\\ & =& \frac{{\sec }^2\theta -{\tan }^2\theta }{\sec \theta -\tan \theta }\\ & =& \frac{1}{\sec \theta -\tan \theta } \left(\because 1+{\tan }^2\theta ={\sec }^2\theta \right)\end{array}$

For 0 ≤ θ < 90°, secθ + tanθ and secθ – tanθ are reciprocal of each other.
Thus, Statement-1 is true.

So, Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Hence, the correct answer is option (b).

Answer:

Statement-2 (Reason): cos²θ + sin²θ = 1.
According to trigonometric identities cos²θ + sin²θ = 1.

Thus, Statement-2 is true.

Statement-1 (Assertion): If x = acosθ and y = bsinθ, then b²x² + a²y² = a²b².
Given, x = acosθ and y = bsinθ.

$\begin{array}{rcl}{b}^2{x}^2+a^2{y}^2& =& b^2\times a^2{\cos }^2\theta +a^2\times b^2{\sin }^2\theta \\ & =& a^2{b}^2\left({\sin }^2\theta +{\cos }^2\theta \right) \left(From Statement‐1\right)\\ & =& a^2{b}^2\end{array}$

Thus, Statement-1 is true.

So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

Answer:

Statement-2 (Reason): For any $x>0, x+\frac{1}{x}\ge 2$.
For any x > 0.
Apply AM ≥ GM on $x \mathrm{and} \frac{1}{x}$,

$$\begin{gathered} \frac{x+{\displaystyle \frac{1}{x}}}{2}\ge \sqrt{x\times \frac{1}{x}} \\ \Rightarrow x+\frac{1}{x}\ge 2 \end{gathered}$$

Thus, statement-2 is true.

Statement-1 (Assertion): For 0 < θ ≤ 90°, cosec²θ​ + sin²θ ≥ 2.
In Statement-1, putting x = sin² θ
$$\begin{gathered} \Rightarrow {\sin }^2\theta +\frac{1}{{\sin }^2\theta }\ge 2 \\ \Rightarrow {\sin }^2\theta +{\mathrm{cosec}}^2\theta \ge 2 \end{gathered}$$

Thus, statement-1 is true.

So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).