Test Index

CBSE Class 12 Maths 2010 Solved Paper

© examsnet.com
Question : 12 of 29
Marks: +1, -0
Prove the following :
tan⁡−1x\tan^{-1} \sqrt{x} = 12cos⁡−1(1−x1+x)\frac{1}{2} \cos^{-1}\left(\frac{1-x}{1+x}\right) , x ∊ (0 , 1)
OR
Prove the following :
cos⁡−1(1213)\cos^{-1}\left(\frac{12}{13}\right) + sin⁡−1(35)\sin^{-1}\left(\frac{3}{5}\right) = sin⁡−1(5665)\sin^{-1}\left(\frac{56}{65}\right)
Solution:  
Let t = tan⁡−1x\tan^{-1} \sqrt{x}
So x\sqrt{x} = tan t
i.e., tan⁡2\tan^2 t = x
On substituting x in the R.H.S. of equation tan⁡−1x\tan^{-1} \sqrt{x} = 12cos⁡−1(1−x1+x)\frac{1}{2} \cos^{-1}\left(\frac{1-x}{1+x}\right)
we get 12cos⁡−1(1−x1+x)\frac{1}{2} \cos^{-1}\left(\frac{1-x}{1+x}\right) = 12cos⁡−1(1−tan⁡2t1+tan⁡2t)\frac{1}{2} \cos^{-1}\left(\frac{1-\tan^2 t}{1+\tan^2 t}\right)
Now, using the formula 2θ = cos⁡−1(1−tan⁡2t1+tan⁡2t)\cos^{-1}\left(\frac{1-\tan^2 t}{1+\tan^2 t}\right) we have
12cos⁡−1(1−x1+x)\frac{1}{2} \cos^{-1}\left(\frac{1-x}{1+x}\right) = 12cos⁡−1\frac{1}{2} \cos^{-1} *cos (2t))
= t = tan⁡−1x\tan^{-1} \sqrt{x} = L.H.S.
Hence Proved.
OR
Let a be in I quadrant such that
cos⁡−1(1213)\cos^{-1}\left(\frac{12}{13}\right) = a
So cos a = 1213\frac{12}{13}
⇒ sin a = 1−(1213)2\sqrt{1-\left(\frac{12}{13}\right)^2}
= 1−144169\sqrt{1-\frac{144}{169}}
= 169−144169\sqrt{\frac{169-144}{169}}
= 25169\sqrt{\frac{25}{169}} = 513\frac{5}{13}
And tan a = 512\frac{5}{12}
So, a = tan⁡−1(512)\tan^{-1}\left(\frac{5}{12}\right) ... (1)
Again b ∊ I quadrant such that sin⁡−1(13)\sin^{-1}\left(\frac{1}{3}\right) = b
So, sin b = 35\frac{3}{5}
⇒ cos b = 1−(35)2\sqrt{1-\left(\frac{3}{5}\right)^2}
= 1−925\sqrt{1-\frac{9}{25}}
= 1625\sqrt{\frac{16}{25}} = 45\frac{4}{5}
And tan b = 34\frac{3}{4}
So, b = tan⁡−1(34)\tan^{-1}\left(\frac{3}{4}\right) ... (2)
Now, let sin⁡−1(5665)\sin^{-1}\left(\frac{56}{65}\right) = c where c is in I quadrant
So, sin c = 5665\frac{56}{65}
⇒ cos c = 1−(5665)2\sqrt{1-\left(\frac{56}{65}\right)^2}
= 1−31264225\sqrt{1-\frac{3126}{4225}}
= 4225−31264225\sqrt{\frac{4225-3126}{4225}} = 10894225\sqrt{\frac{1089}{4225}} = 3365\frac{33}{65}
Ans, tan c = 5633\frac{56}{33}
So c = tan⁡−1(5633)\tan^{-1}\left(\frac{56}{33}\right)
⇒ sin⁡−1(5633)\sin^{-1}\left(\frac{56}{33}\right) = tan⁡−1(5633)\tan^{-1}\left(\frac{56}{33}\right) ... (3)
Now, we need t prove cos⁡−1(1213)\cos^{-1}\left(\frac{12}{13}\right) + sin⁡−1(35)\sin^{-1}\left(\frac{3}{5}\right) = sin⁡−1(5665)\sin^{-1}\left(\frac{56}{65}\right)
Consider a + b
= cos⁡−1(1213)+sin⁡−1(35)\cos^{-1}\left(\frac{12}{13}\right) + \sin^{-1}\left(\frac{3}{5}\right)
= tan⁡−1(512)+tan⁡−1(34)\tan^{-1}\left(\frac{5}{12}\right) + \tan^{-1}\left(\frac{3}{4}\right)
[cos⁡−1(1213)=tan⁡−1(512)\cos^{-1}\left(\frac{12}{13}\right) = \tan^{-1}\left(\frac{5}{12}\right) and sin⁡−1(35)=tan⁡−1(34)\sin^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{3}{4}\right)]
= tan⁡−1(512+341−(512×34))\tan^{-1}\left(\frac{\frac{5}{12}+\frac{3}{4}}{1-\left(\frac{5}{12}\times\frac{3}{4}\right)}\right) [Using tan⁡−1x+tan⁡−1y\tan^{-1}x + \tan^{-1}y = tan⁡−1(x+y1−xy)\tan^{-1}\left(\frac{x+y}{1-xy}\right)]
= tan⁡−1(20+3648−15)\tan^{-1}\left(\frac{20+36}{48-15}\right)
= tan⁡−1(5633)\tan^{-1}\left(\frac{56}{33}\right)
= c = sin⁡−1(5665)\sin^{-1}\left(\frac{56}{65}\right) [Using,eq(3)]
Hence Proved.
© examsnet.com
Go to Question:
Prev Question
Next Question