If cos ^{-1} sqrt{p}+cos ^{-1} sqrt{1-p}+cos | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given equation: $\cos ^{-1} \sqrt{p}+\cos ^{-1} \sqrt{1-p}+\cos ^{-1} \sqrt{1-q}=\frac{3 \pi}{4}$ Let $t=\cos ^{-1} \sqrt{p}$. Then $\sqrt{p}=\cos

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) Given equation: $\cos ^{-1} \sqrt{p}+\cos ^{-1} \sqrt{1-p}+\cos ^{-1} \sqrt{1-q}=\frac{3 \pi}{4}$ Let $t=\cos ^{-1} \sqrt{p}$. Then $\sqrt{p}=\cos t$,so $p=\cos ^2 t$. Thus,$1-p=1-\cos ^2 t=\sin ^2 t$,which implies $\sqrt{1-p}=\sin t$. Therefore,$t=\sin ^{-1} \sqrt{1-p}$. Since $t=\cos ^{-1} \sqrt{p}$,we have $\cos ^{-1} \sqrt{p}=\sin ^{-1} \sqrt{1-p}$. Substituting this into the given equation: $\sin ^{-1} \sqrt{1-p}+\cos ^{-1} \sqrt{1-p}+\cos ^{-1} \sqrt{1-q}=\frac{3 \pi}{4}$ Using the identity $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$,we get: $\frac{\pi}{2}+\cos ^{-1} \sqrt{1-q}=\frac{3 \pi}{4}$ $\cos ^{-1} \sqrt{1-q}=\frac{3 \pi}{4}-\frac{\pi}{2}=\frac{\pi}{4}$ $\sqrt{1-q}=\cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$ Squaring both sides: $1-q=\frac{1}{2}$ $q=1-\frac{1}{2}=\frac{1}{2}$

If $\cos ^{-1} \sqrt{p}+\cos ^{-1} \sqrt{1-p}+\cos ^{-1} \sqrt{1-q}=\frac{3 \pi}{4}$,then the value of $q$ is:

Similar Questions

$\begin{aligned} & 2 \sin ^{-1} x+\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)+3 \cos ^{-1} x \\ & -\cos ^{-1}\left(4 x^3-3 x\right) \text { is equal to }\end{aligned}$

If $y = \tan^{-1}\sqrt{\frac{1 + \cos x}{1 - \cos x}}$,then $\frac{dy}{dx}$ is equal to

$\cos ^{-1}\left(\frac{-1}{2}\right)-2 \sin ^{-1}\left(\frac{1}{2}\right)+3 \cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)-4 \tan ^{-1}(-1)$ equals

Solve for $x$: $\tan^{-1}\left(\frac{1-x}{1+x}\right) = \frac{1}{2} \tan^{-1} x$,where $x > 0$.

If $x=\operatorname{cosec}(\tan ^{-1}(\cos (\cot ^{-1}(\sec (\sin ^{-1} a)))))$,where $a \in [0, 1]$,then which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module