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Question of 108

Q.Prove that: tan⁡12[sin⁡−1(2x1+x2)+cos⁡−1(1−y21+x2)]=x+y1−xy\tan\dfrac{1}{2}\left[\sin^{-1}\left(\dfrac{2x}{1+x^2}\right) + \cos^{-1}\left(\dfrac{1-y^2}{1+x^2}\right)\right] = \dfrac{x+y}{1-xy}.

Manipur CohsemCOHSEM Manipur Higher Secondary Board 2025Subjective· 2mImportance★★★★★
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The second term's denominator is corrected from the printed 1+x21+x^2 to 1+y21+y^2 (the standard identity, and the only reading that produces the stated RHS); then use the double-angle inverse-trig identities and the tan-addition formula.

Note on the stem: the second term is printed as cos⁡−1(1−y21+x2)\cos^{-1}\left(\dfrac{1-y^2}{1+x^2}\right), mixing xx and yy in one fraction. The standard NCERT identity (and the only reading consistent with the given RHS x+y1−xy\dfrac{x+y}{1-xy}) uses cos⁡−1(1−y21+y2)\cos^{-1}\left(\dfrac{1-y^2}{1+y^2}\right). Solving with this corrected reading.

Standard identities (for x,y≥0x,y\ge0, within the valid ranges):

sin⁡−1(2x1+x2)=2tan⁡−1x\sin^{-1}\left(\dfrac{2x}{1+x^2}\right) = 2\tan^{-1}x

cos⁡−1(1−y21+y2)=2tan⁡−1y\cos^{-1}\left(\dfrac{1-y^2}{1+y^2}\right) = 2\tan^{-1}y

Substituting into the LHS:

tan⁡12[2tan⁡−1x+2tan⁡−1y]=tan⁡[tan⁡−1x+tan⁡−1y]\tan\dfrac12\left[2\tan^{-1}x+2\tan^{-1}y\right] = \tan\left[\tan^{-1}x+\tan^{-1}y\right]

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