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Trigonometric Ratios of Complementary Angles - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

Complementary Angles is considered one the most difficult concept.
13 Questions around this concept.

Solve by difficulty

If $\cos ^{-1}x-\cos ^{-1}\frac{y}{2}=\alpha ,$ then $4x^{2}-4xy\cos \alpha +y^{2}$ is equal to:

A

$4$

B

$2\sin 2\alpha$

C

$-4\sin ^{2}\alpha$

D

$4\sin ^{2}\alpha$

Let $\tan ^{-1}y=\tan ^{-1}x\, +\, \tan ^{-1}\left ( \frac{2x}{1-x^{2}} \right ),$ where $\left | x \right |< \frac{1}{\sqrt{3}}.$

Then the value of $y$ is :

A

$\frac{3x-x^{3}}{1-3x^{2}}$

B

$\frac{3x+x^{3}}{1-3x^{2}}$

C

$\frac{3x-x^{3}}{1+3x^{2}}$

D

$\frac{3x+x^{3}}{1+3x^{2}}$

Concepts Covered - 1

Complementary Angles

Complementary Angles

$\\1.\;\;\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}, \text { for all } x \in[-1,1] \\\\2.\;\; \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}, \text { for all } x \in R \\\\3.\;\; \sec ^{-1} x+\cos \operatorname{ec}^{-1} x=\frac{\pi}{2} \text { for all } x \in(-\infty,-1] \cup[1, \infty)$

Proof:

1.

$\\\text { Let } \sin ^{-1} x=\theta\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(i) \\ \text { where } \theta \in[-\pi / 2 , \pi / 2]\\\\ \Rightarrow \quad\quad\quad-\frac{\pi}{2} \leq-\theta \leq \frac{\pi}{2} \\\\ \Rightarrow \quad\quad\quad 0 \leq \frac{\pi}{2}-\theta \leq\pi \\\\\Rightarrow \quad\quad\quad \frac{\pi}{2}-\theta \in[0,\pi] \\\\ \text { Now, } \sin ^{-1} x=\theta\\\text { Or, } \;\;\;\;\;\;\;\;x=\sin\theta\\\\ \Rightarrow \quad\quad\quad x=\cos \left(\frac{\pi}{2}-\theta\right)\\\\\mathrm{\Rightarrow \quad\quad\quad \cos ^{-1} x=\frac{\pi}{2}-\theta\;\;\;\;\;\;\;\;\;[\because x \in[-1,1] \text { and }(\pi / 2-\theta) \in[0, \pi]]}\\\\\mathrm{\Rightarrow \quad\quad\quad \theta+\cos ^{-1} x=\frac{\pi}{2}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(ii)\\\\\text { From Eqs.(i) and (ii), we get } \sin ^{-1} x+\cos ^{-1} x=\pi / 2 . \text { Similarly, we get the other results. }$

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Complementary Angles

Mathematics for Joint Entrance Examination JEE (Advanced) : Trigonometry

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