If $sin x + {sin}^{2} x = 1,$ then the value of ${cos}^{12} x + 3 {cos}^{10} x + 3 {cos}^{8} x + {cos}^{6} x - 2$ is equal to

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $- 1$
$s i n x + {s i n}^{2} x = 1 \Rightarrow s i n x = 1 - {s i n}^{2} x \Rightarrow s i n x = {c o s}^{2} x$ ...(1)
${c o s}^{12} x + 3 {c o s}^{10} x + 3 {c o s}^{8} x + {c o s}^{6} x - 2$
Substituting values from (1), we get
${s i n}^{6} x + 3 {s i n}^{5} x + 3 {s i n}^{4} x + {s i n}^{3} x - 2 = {({s i n}^{2} x + s i n x)}^{3} - 2 = 1 - 2 = - 1$

Suggest Corrections

Similar questions

sin x + {sin}^{2} x = 1

{cos}^{12} x + 3 {cos}^{10} x + 3 {cos}^{8} x + {cos}^{6} x - 2

If sinx + $s i n^{2} x$ = 1 Then the value of $c o s^{12} x + 3 c o s^{10} x + 3 c o s^{8} x + c o s^{6} x$ is equal to

$I f s i n x + s i n^{2} x = 1, t h e n t h e v a l u e o f c o s^{12} x + 3 c o s^{10} x + 3 c o s^{8} x + c o s^{6} x - 1 i s e q u a l t o$

sin x + {sin}^{2} x = 1

{cos}^{12} x + 3 {cos}^{10} x + 3 {cos}^{8} x + {cos}^{6} x + 2 {cos}^{4} x + {cos}^{2} x - 2 =

s i n x + s i n^{2} x = 1

c o s^{2} x + c o s^{4} x

Join BYJU'S Learning Program