Question

if sin theta + cosec theta = 2 then the value of sin 2016 theta + cosec 2016 theta = ?​

Answer 1

Sine and cosecant trigonometric ratios are reciprocals to each other, aren't they?

Let  sin θ  be  x.  So  csc θ  will be  1/x.

$\begin{aligned}&\sin\theta+\csc\theta=2\\ \\ \Longrightarrow\ \ &x+\dfrac{1}{x}=2\end{aligned}$

And it is always true that,

$\text{If}\ \ \ x+\dfrac{1}{x}=2,\ \ \ \text{then}\ \ \ x^n+\dfrac{1}{x^n}=2$

(Proof of this concept is given at the end.)

So, taking n = 2016, we get that,

$\begin{aligned}&x^{2016}+\dfrac{1}{x^{2016}}&=&\ \ 2\\ \\ \Longrightarrow\ \ &x^{2016}+\dfrac{1^{2016}}{x^{2016}}&=&\ \ 2\\ \\ \Longrightarrow\ \ &x^{2016}+\left(\dfrac{1}{x}\right)^{2016}&=&\ \ 2\end{aligned}$

Thus,

$\begin{aligned}\\ \\ \Longrightarrow\ \ &\sin^{2016}\theta+\csc^{2016}\theta&=&\ \ \Large \textbf{2}\end{aligned}$

Hence the answer is also 2.

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$\large \textsc{\underline{\underline{PROOF\ OF\ THE\ ABOVE\ CONCEPT}}}$

$\begin{aligned}&x+\dfrac{1}{x}&=&\ \ 2\\ \\ \Longrightarrow\ \ &\frac{x^2}{x}+\frac{1}{x}&=&\ \ 2\\ \\ \Longrightarrow\ \ &\frac{x^2+1}{x}&=&\ \ 2\\ \\ \Longrightarrow\ \ &x^2+1&=&\ \ 2x\\ \\ \Longrightarrow\ \ &x^2-2x+1&=&\ \ 0\\ \\ \Longrightarrow\ \ &(x-1)^2&=&\ \ 0\\ \\ \Longrightarrow\ \ &x-1&=&\ \ 0\\ \\ \Longrightarrow\ \ &x&=&\ \ 1\end{aligned}$

$\begin{aligned}\therefore\ \ \ \ &x^n+\frac{1}{x^n}&=&\ \ 1^n+\frac{1}{1^n}\\ \\ \Longrightarrow\ \ &x^n+\frac{1}{x^n}&=&\ \ 1+\frac{1}{1}\\ \\ \Longrightarrow\ \ &x^n+\frac{1}{x^n}&=&\ \ 1+1\\ \\ \Longrightarrow\ \ &x^n+\frac{1}{x^n}&=&\ \ 2\end{aligned}\\ \\ \\ \\ \textsf{Hence Proved!}$