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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.

  • 24 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:

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 :

If $\operatorname{Sin}^{-1} x=\frac{\pi}{5}$ for some $x \epsilon(-1,1)$ then value of $\cos ^{-1} x$ is

The value of $Cot^{-1}\frac{3}{4}+Sin^{-1}\frac{5}{13}$ is

Number of solutions of  $3\cos^{-1}x= \frac{\pi}{2}-\sin^{-1}x$  is

If $\cos\left ( sin^{-1}\frac{2}{5} + cos^{-1}x \right )=0,$ then x is equal to

 

Concepts Covered - 1

Complementary Angles

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

Proof:
1.

Let $\sin ^{-1} x=\theta$
where $\theta \in[-\pi / 2, \pi / 2]$

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

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Complementary Angles

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