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CBSE Class 12 Math 2011 Solved Paper

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Question : 12 of 29
Marks: +1, -0
Prove the following :
cot⁡−1[1+sin⁡x+1−sin⁡x1+sin⁡x−1−sin⁡x]\cot^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right] = x2\frac{x}{2} , x ∊ (0,π2)\left(0,\frac{\pi}{2}\right)
OR
Find the value of tan⁡−1(xy)\tan^{-1}\left(\frac{x}{y}\right) - tan⁡−1(x−yx+y)\tan^{-1}\left(\frac{x-y}{x+y}\right)
Solution:  
cot⁡−1[1+sin⁡x+1−sin⁡x1+sin⁡x−1−sin⁡x]\cot^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]
=
cot⁡−1[sin⁡2x2+cos⁡2x2+sin⁡2(x2)+sin⁡2x2+cos⁡2x2−sin⁡2(x2)sin⁡2x2+cos⁡2x2+sin⁡2(x2)−sin⁡2x2+cos⁡2x2−sin⁡2(x2)]\cot^{-1}\left[\frac{\sqrt{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}+\sin 2\left(\frac{x}{2}\right)}+\sqrt{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}-\sin 2\left(\frac{x}{2}\right)}}{\sqrt{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}+\sin 2\left(\frac{x}{2}\right)}-\sqrt{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}-\sin 2\left(\frac{x}{2}\right)}}\right]
[Since, sin⁡2\sin^2 A + cos⁡2\cos^2 A = 1]
=
cot⁡−1[sin⁡2x2+cos⁡2x2+2sin⁡x2cos⁡x2+sin⁡2x2+cos⁡2x2−2sin⁡x2cos⁡x2sin⁡2x2+cos⁡2x2+2sin⁡x2cos⁡x2−sin⁡2x2+cos⁡2x2−2sin⁡x2cos⁡x2]\cot^{-1}\left[\frac{\sqrt{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}}+\sqrt{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}-2\sin\frac{x}{2}\cos\frac{x}{2}}}{\sqrt{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}}-\sqrt{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}-2\sin\frac{x}{2}\cos\frac{x}{2}}}\right]
[Since, sin2A = 2 sinA cosA]
=
cot⁡−1[(cos⁡x2+sin⁡x2)2+(cos⁡x2−sin⁡x2)2(cos⁡x2+sin⁡x2)2−(cos⁡x2−sin⁡x2)2]\cot^{-1}\left[\frac{\sqrt{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^2}+\sqrt{\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)^2}}{\sqrt{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^2}-\sqrt{\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)^2}}\right]
= cot⁡−1(2cos⁡x22sin⁡x2)\cot^{-1}\left(\frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}}\right) = cot⁡−1(cot⁡x2)\cot^{-1}\left(\cot\frac{x}{2}\right)
= x2\frac{x}{2}
Hence proved.
OR
tan⁡−1(xy)\tan^{-1}\left(\frac{x}{y}\right) - tan⁡−1(x−yx+y)\tan^{-1}\left(\frac{x-y}{x+y}\right)
tan⁡−1(xy)\tan^{-1}\left(\frac{x}{y}\right) - tan⁡−1(xy−1xy+1)\tan^{-1}\left(\frac{\frac{x}{y}-1}{\frac{x}{y}+1}\right)
= tan⁡−1(xy)\tan^{-1}\left(\frac{x}{y}\right) - tan⁡−1(xy−11+xy)\tan^{-1}\left(\frac{\frac{x}{y}-1}{1+\frac{x}{y}}\right)
tan⁡−1(xy)\tan^{-1}\left(\frac{x}{y}\right) - [tan⁡−1(xy)−tan⁡−1(1)]\left[\tan^{-1}\left(\frac{x}{y}\right)-\tan^{-1}(1)\right] [Since tan⁡−1\tan^{-1} a - tan⁡−1\tan^{-1} b = tan⁡−1(a−b1+ab)\tan^{-1}\left(\frac{a-b}{1+ab}\right)]
tan⁡−1(xy)\tan^{-1}\left(\frac{x}{y}\right) - tan⁡−1(xy)\tan^{-1}\left(\frac{x}{y}\right) + tan⁡−1(1)\tan^{-1}(1)
= tan⁡−1(1)\tan^{-1}(1) = π4\frac{\pi}{4}
Thus tan⁡−1(xy)\tan^{-1}\left(\frac{x}{y}\right) - tan⁡−1(x−yx+y)\tan^{-1}\left(\frac{x-y}{x+y}\right) = π4\frac{\pi}{4}
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