Question

if
$sin(a) + cosec(a) = 2$
then find the value of
$sin {8} (a) + cosec {8} (a)$

​

Answer 1

Answer:

sinθ+cosecθ=2

We know that, cosecθ=

sinθ

1

sinθ+

sinθ

1

=2

sin

2

θ−2sinθ+1=0

(sinθ−1)

2

=0

(sinθ−1)=0

sinθ=1

sinθ=sin90

θ=90

Now, find the value.

sin

8

θ+cosec

8

θ=(sinθ)

8

+(cosecθ)

8

=(sin90)

8

+(cosec90)

8

=(1)

8

+(1)

8

=1+1

=2

sin

8

θ+cosec

8

θ=2

Hence, the required result is found.

Answer 2

$\huge \rm {\underbrace{\underline{Answer:-}}}$

$\sf \red {\underline{\underline{Provided\: that:}}}$

$\to \tt {Sin\:a+cosec\:a=2}$

$\sf \blue {\underline{\underline{To\: Determine:}}}$

$\to \tt {sin^{8} \:a + cosec^{8} \:a=?}$

➻Firstly,we need to calculate the value of 'a'

$\sf \pink {\underline{\underline{We\: know:}}}$

$\large \to \tt {cosec\:a=\frac{1}{sin\:a}}$

➻so we can write,

$\large \to \tt {Sin\:a+cosec\:a=2}$

➻As ,

$\to \tt {Sin\:a+\frac{1}{sin\:a}=2}$

➻ Taking L.C.M,

$\large \to \tt {\frac{sin\:a\times sin\:a+1}{sin(a)}=2}$

$\large \to \tt {\frac{sin^{2}\:a+1}{sin\:a}=2}$

$\to \tt {sin^{2}\:a+1=2sin\:a}$

$\to \tt {sin^{2}\:a-2sin\:a+1=0}$

➻It is in the form of $\tt {a^{2}-2ab+b^{2}=(a-b)^{2}}$

➻where ,

$\to \tt {a=sin\:a}$

$\to \tt {b=1}$

➻so it can be written as $\tt {(a-b)^{2}}$

$\to \tt {(sin\:a-1)^{2}=0}$

$\to \tt {sin\:a=1}$

$\to \tt {sin\:a=sin90°}$ [∵sin90°=1]

$\to \tt {\cancel{sin}\:a=\cancel{sin}90°}$

$\implies \green {\fbox{a=90°}}$

➻As the value of 'a' is determined,

$\to \tt {sin^{8} \:a + cosec^{8} \:a}$

$\to \tt {(sin \:a )^{8}+ (cosec \:a)^{8}}$

➻Substituting the value of 'a'

$\to \tt {(sin90° )^{8}+ (cosec 90°)^{8}}$

$\to \tt {(1)^{8}+ (1)^{8}}$

[∵sin90°=1 and cosec90°=1]

$\to \tt {1+1}$

$\large \implies \green {\fbox{2}}$

$\sf \purple {\underline{\underline{Henceforth,}}}$

➻The required Answer is "2"