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  • If sin inverse 2a divided 1plus a square plus cos inverse 1 minus a square divided 1 plus a square equal to tan inverse 2x divided 1 minus x square where a, x belong to 0,1 then the vale of x is A. 0 B.a divided 2 C. a D. 2a divided 1 minus a square

If \sin^{-1}\frac{2a}{1+a^{2}}+\cos ^{-1}\frac{1-a^{2}}{1+a^{2}}=tan^{-1}\frac{2x}{1-x^{2}} where a, x\epsilon \left [ 0,1 \right ] then the value of x is

A. 0

B. a/2

C. a

D. \frac{2a}{1-a^{2}}

Answers (1)

Answer:(D)

We have 

sin^{-1}\frac{2a}{1+a^{2}}+cos^{-1}\frac{1-a^{2}}{1+a^{2}}=\tan^{-1}\frac{2x}{1-x^{2}}

we know that

2 \tan ^{-1}p=\sin^{-1}\frac{2p}{1+p^{2}}.........(1)

Also,2 \tan^{-1}p=\cos^{-1}\frac{1-p^{2}}{1+p^{2}}.........(2)

Also,2 \tan^{-1}p=\tan^{-1}\frac{2p}{1-p^{2}}.........(3)

From (1) and (2) we have,

L.H.S-

\sin^{-1}\frac{2a}{1+a^{2}}+\cos^{-1}\frac{1-a^{2}}{1+a^{2}}=2\tan^{-1}a+2\tan^{-1}a

\Rightarrow \sin^{-1}\frac{2a}{1+a^{2}}+\cos^{-1}\frac{1-a^{2}}{1+a^{2}}=4\tan^{-1}a

From (3) R.H.S

\tan^{-1}\frac{2x}{1-x^{2}}=2\tan^{-1}x

So, we have 4 tan-1 a = 2 tan-1 x

⇒ 2 tan-1 a = tan-1 x

But from (3) 2\tan^{-1}a= \tan^{-1}\frac{2a}{1-a^{2}}

So \tan^{-1}\frac{2a}{1-a^{2}}=\tan^{-1}x

x=\frac{2a}{1-a^{2}}

 

 

 

Posted by

infoexpert24

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