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

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Solution

The correct option is A
0

From sin $x + s i n^{2} x = 1$ , we get sin $x = c o s^{2} x$ , Now, the given expression is equal to $c o s^{6} x (c o s^{6} x + 3 c o s^{4} x + 3 c o s^{2} x + 1) - 1 = c o s^{6} x (c o s^{2} x + 1) 3 - 1 = s i n^{3} x (s i n x + 1)^{3} - 1 = (s i n^{2} x + s i n x)^{3} - 1 = 1 - 1 = 0$

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Similar questions

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

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

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

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 =

sin x = \frac{2 t}{1 + t^{2}}, tan y = \frac{2 t}{1 - t^{2}}

\frac{d y}{d x}

Δ (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{x} & s i n 2 x & t a n x^{2} l n (1 + x) & c o s x & s i n x c o s x^{2} & e^{x} - 1 & s i n x^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

= A + B x + C x^{2} + \dots

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