3.Trigonometrical Ratios, Functions and Identities
hard

किसी $\theta \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$ के लिये, व्यंजक $3(\sin \theta-\cos \theta)^{4}+6(\sin \theta+\cos \theta)^{2}+4 \sin ^{6} \theta$ होगा

A

$13 - 4\,{\cos ^2}\,\theta \, + 6\,{\sin ^2}\,\theta \,{\cos ^2}\,\theta $

B

$13 - 4\,{\cos ^6}\,\theta \,$

C

$13 - 4\,{\cos ^2}\,\theta \, + 6\,\,{\cos ^4}\,\theta $

D

$13 - 4\,{\cos ^4}\,\theta \, + 2\,{\sin ^2}\,\theta \,{\cos ^2}\,\theta $

(JEE MAIN-2019)

Solution

$3\,{(1 – \sin 2\theta )^2}\, + \,6(1 + \sin 2\theta )\, + \,4\,{\sin ^6}\theta $

$ = 3\,(1 – 2\sin \,2\theta  + {\sin ^2}2\theta ) + \,6 + 6\sin 2\theta  + \,4\,{\sin ^6}\theta $

$ = \,9 + 3{\sin ^2}2\theta  + 4\,{\sin ^6}\theta $

$ = \,9 + 12{\sin ^2}\theta {\cos ^2}\theta  + 4\,{(1 – {\cos ^2}\theta )^3}$

$ = \,9 + 12(1 – {\cos ^2}\theta ){\cos ^2}\theta  + 4\,(1 – 3{\cos ^2}\theta  + 3{\cos ^4}\theta  – {\cos ^6}\theta )$

$ = \,13 + 12{\cos ^2}\theta  – 12{\cos ^4}\theta  – 12{\cos ^2}\theta  + 12{\cos ^4}\theta  – 4{\cos ^6}\theta $

$ = \,13 – 4{\cos ^6}\theta $

Standard 11
Mathematics

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