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Q.Prove that cos⁡−16365+2tan⁡−115=sin⁡−135\cos^{-1}\dfrac{63}{65} + 2\tan^{-1}\dfrac{1}{5} = \sin^{-1}\dfrac{3}{5}. OR Solve the equation tan⁡−13x+tan⁡−12x=π4\tan^{-1}3x + \tan^{-1}2x = \dfrac{\pi}{4}.

Rajasthan RbseRajasthan Board Senior Secondary Examination 2019Subjective· 3mImportance★★★★★
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Convert 2tan⁡−1(1/5)2\tan^{-1}(1/5) to a single arctan, combine with cos⁡−1(63/65)\cos^{-1}(63/65) using the tangent addition formula, and show the result equals sin⁡−1(3/5)\sin^{-1}(3/5). (OR: use the tangent-sum formula to solve for xx.)

Part 1: Prove cos⁡−16365+2tan⁡−115=sin⁡−135\cos^{-1}\dfrac{63}{65}+2\tan^{-1}\dfrac15 = \sin^{-1}\dfrac35

First, 2tan⁡−1152\tan^{-1}\dfrac15: using tan⁡2α=2tan⁡α1−tan⁡2α\tan2\alpha = \dfrac{2\tan\alpha}{1-\tan^2\alpha} with tan⁡α=15\tan\alpha=\dfrac15:

tan⁡2α=2/51−1/25=2/524/25=512\tan2\alpha = \dfrac{2/5}{1-1/25} = \dfrac{2/5}{24/25} = \dfrac{5}{12}, so 2tan⁡−115=tan⁡−15122\tan^{-1}\dfrac15 = \tan^{-1}\dfrac{5}{12}.

Next, let β=cos⁡−16365\beta=\cos^{-1}\dfrac{63}{65}, so cos⁡β=6365\cos\beta=\dfrac{63}{65} and sin⁡β=1−(6365)2=1665\sin\beta=\sqrt{1-\left(\frac{63}{65}\right)^2}=\dfrac{16}{65} (using 652−632=(65−63)(65+63)=2×128=25665^2-63^2=(65-63)(65+63)=2\times128=256), so tan⁡β=1663\tan\beta=\dfrac{16}{63}.

Now add: tan⁡(β+tan⁡−1512)=tan⁡β+5121−tan⁡β⋅512=1663+5121−1663⋅512\tan\left(\beta+\tan^{-1}\tfrac{5}{12}\right) = \dfrac{\tan\beta+\frac{5}{12}}{1-\tan\beta\cdot\frac{5}{12}} = \dfrac{\frac{16}{63}+\frac{5}{12}}{1-\frac{16}{63}\cdot\frac{5}{12}}

Over a common denominator 756756: numerator =192+315756=507756=\dfrac{192+315}{756}=\dfrac{507}{756}; the denominator (of the big fraction) =1−80756=676756=1-\dfrac{80}{756}=\dfrac{676}{756}.

So the sum's tangent =507676=34=\dfrac{507}{676}=\dfrac34 (dividing both by 169169).

An angle with tan⁡=34\tan=\dfrac34 (in the first quadrant, as both original angles are acute/positive) has sin⁡=35\sin=\dfrac35 (3-4-5 right triangle). Hence β+2tan⁡−115=sin⁡−135\beta+2\tan^{-1}\tfrac15 = \sin^{-1}\dfrac35, i.e. LHS == RHS. Proved.

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