If cot ⁡ ( cos^-1 ⁡x )= sec ⁡ ( tan^-1 ⁡ ( a b^2-a^2 ) ) ba, then x=

Question

If cot ⁡ ( cos^-1 ⁡x )= sec ⁡ ( tan^-1 ⁡ ( a b^2-a^2 ) ) b>a, then x=

Accepted Answer

Correct answer: A. (a) Given that, cot ⁡ ( cos^-1 ⁡x )= sec ⁡ ( tan^-1 ⁡ ( a b^2-a^2 ) ) Since , cos^-1 ⁡x= cot^-1 ⁡ ( x 1-x^2 ) and tan^-1 ⁡x= sec^-1 ⁡ ( 1+x^2 ) ⇒ cot ⁡ ( cot^-1 ⁡ x 1-x^2 )= sec ⁡ ( sec^-1 ⁡ 1+ ( a b^2-a^2 )^2 ) ⇒ x 1-x^2= b^2-a^2+a^2b^2-a^2 ⇒ b 1-x^2= b b^2-a^2 ⇒ b 1-x^2= b b^2-a^2 On squaring both the sides, we get x^21-x^2= b^2b^2-a^2 x^2b^2-x^2a^2=b^2-b^2x^2 x^2b^2+b^2x^2-x^2a^2=b^2 2x^2b^2-x^2a^2=b^2 ⇒x^2(2b^2-a^2)=b^2 ⇒x= b 2b^2-a^2

$If cot (cos^{- 1} x) = sec (tan^{- 1} (\frac{a}{b ^{2} - a ^{2}})) b > a, then x =$

Solution

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