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  • If<img alt="\sin ^{-1}\left(\frac{2 a}{1+a^{2}}\right)+\sin ^{-1}\left(\frac{2 b}{1+b^{2}}\right)=2 \tan ^{-1} x" src="http://entrancecorner.oncodecogs.com/gif.latex?%5Csin%20%5E%7B-1%7D%5Cleft%28%

If\sin ^{-1}\left(\frac{2 a}{1+a^{2}}\right)+\sin ^{-1}\left(\frac{2 b}{1+b^{2}}\right)=2 \tan ^{-1} x then x= ?

Option: 1

{\frac{a-b}{1+a b}}


Option: 2

{\frac{b}{1+a b}}


Option: 3

\\ {\frac{b}{1-a b}}


Option: 4

{\frac{a+b}{1-a b}}


Answers (1)

best_answer

\sin ^{-1}(\frac{2a}{1+a^{2}}+\sin ^{-1}(\frac{2 b}{1+b^{2}})\\ \text {We can observe here that if we put } a=\tan \alpha \text { and } b=\tan \beta \text { it will become formula of } \sin 2\alpha \ and\ \sin 2 \beta \\ \sin ^{-1}(\frac{2 a}{1+a^{2}}+\sin ^{-1}(\frac{2 b}{1+b^{2}})=2 \tan ^{-1} x \\ \sin ^{-1}(\frac{2 \tan \alpha}{1+\tan ^{2} \alpha})+\sin ^{-1}(\frac{2 \tan \beta}{1+\tan ^{2} \beta})=2 \tan ^{-1} x \\ \sin ^{-1} \sin (2 \alpha)+\sin ^{-1} \sin (2 \beta)=2 \tan ^{-1} x \\ 2(\alpha+\beta)= 2 \tan ^{-1} x \\ x=\tan (\alpha +\beta) \\ x= \frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \\ x=\frac{a+b}{1-a b}

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vishal kumar

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