Question

If sin 0 + cosec 0 = 2, then sin? 0 + cosec? @ is equal to :​

Answer 1

Answer:

ANSWER

Given sinθ+cosecθ=2

⟹sinθ+

sinθ

1

=2

squaring on both sides

(sinθ+

sinθ

1

)

2

=4

⟹sin

2

θ+

sin

2

θ

1

+2(sinθ)(

sinθ

1

)=4

⟹sin

2

θ+

sin

2

θ

1

+2=4

⟹sin

2

θ+

sin

2

θ

1

=2

⟹sin

2

θ+cosec

2

θ=2

Answer 2

Answer:

Option (A) 2 is the correct answer.

Explanation:

${\boxed{\boxed{\begin{array}{cc}\bf \: \to \:Given : \\ \\ \rm \: sin \theta + cosec\theta = 2 \\ \\ \blue{ \underline{ \pink{\text{We \: have \: to \: find : }}}} \\ \\ \sf \: The \: value \: of \: \boxed{ \rm \: (sin {}^{8} \theta+ cosec {}^{8}\theta )}\end{array}}}}$

${\boxed{\boxed{\begin{array}{cc}\red{ \underline{ \rm{ \: Solution\: - {1}^{st} \: step }}} \\ \\ \rm \: sin\theta + cosec\theta = 2 \\ \\ \rm \: \implies \:sin\theta + \frac{1}{sin\theta} = 2 \\ \\ \rm \: \implies \: \frac{sin {}^{2} \theta + 1}{sin\theta} = 2 \\ \\ \rm \: \implies \:sin {}^{2} \theta + 1 = 2 \: sin\theta \\ \\ \rm \: \implies \:sin {}^{2} \theta - 2 \: sin\theta + 1 = 0 \\ \\ \orange{{\boxed{\begin{array}{cc}\bf \: \to \: we \: know \: that : \\ \\ ( {x - y)}^{2} = {x}^{2} - 2xy+ {y}^{2} \end{array}}}} \\ \sf \: apply \: this \\ \\ \rm \: \implies \:( {sin\theta - 1)}^{2} = 0 \\ \\ \rm \: \implies \:sin\theta - 1 = 0 \\ \\ \rm \: \implies \:sin\theta = 1 \\ \\ \pink{\therefore \: \rm \: sin\theta \: = 1 \: \: \: - - - (1)}\end{array}}}}$

${\boxed{\boxed{\begin{array}{cc}\red{ \underline{ \rm \: Solution - {2}^{nd} \: step}} \\ \\ \bf \: now \: \\ \\ \rm \: {sin}^{8}\theta + {cosec}^{8} \theta \\ \\ \rm= {sin}^{8} \theta + \frac{1}{ \rm {sin}^{8}\theta } \\ \\ \rm = {(sin\theta)}^{8} + \frac{1}{ {(sin\theta)}^{8} } \\ \\ = 1 {}^{8} + \frac{1}{1 {}^{8} } \: \: \: \pink{ \{ \sf \: by \: eqn.(1) \}} \\ \\ = 1 + 1 \\ \\ = 2 \\ \\ \orange { \boxed{\therefore \rm \: {sin}^{8}\theta + {cosec}^{8}\theta = 2}} \end{array}}}}$

Short rule:

If sin x + cosec x = 2

Then, sinⁿ x + cosecⁿ x = 2