{sin ^6}θ + {cos ^6}θ + 3{sin ^2}θ {cos ^2}θ =

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3.Trigonometrical Ratios, Functions and Identities

easy

${\sin ^6}\theta + {\cos ^6}\theta + 3{\sin ^2}\theta {\cos ^2}\theta = $

A

$0$

B

$-1$

C

$1$

D

None of these

Solution

(c) ${\sin ^6}\theta + {\cos ^6}\theta + 3\,{\sin ^2}\theta \,{\cos ^2}\theta $

$ = {({\sin ^2}\theta + {\cos ^2}\theta )^3} – 3{\sin ^2}\theta {\cos ^2}\theta + 3{\sin ^2}\theta {\cos ^2}\theta = 1.$

Trick : Put $\theta = {0^o},$

we get the value of expression equal to $1. $

Again put $\theta = {45^o},$ the value remains $1,$ it means that the expression is independent of $\theta$ and is equal to $1.$

Standard 11

Mathematics

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