Question

prove this question. cosec²A+cosec²(90-A)=cosec²A.cosec²(90-A).​

Answer 1

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Answer 2

Observe the angles of which the cosecant values are given.

Here the angles mentioned are A and (90 - A).  Their sum is 90°.

These two angles can be said as the angles of a right triangle except 90°.

Because the other two angles of any right triangle except 90° are in the form x and 90 - x.

We know that the cosecant ratio of an angle is the ratio of the length of the hypotenuse of a right triangle to that of the side opposite to that angle. It's the reciprocal of sine ratio, isn't it?

We have to consider the angle A mentioned in the question.

As we've seen earlier, the side opposite to the angle (90 - A) is the side adjacent to the angle A.

Thus, according to the angle A,

⇒  csc A = hyp / opp

⇒  csc (90 - A) = hyp / adj

Okay, proving...

$\displaystyle \begin{aligned}&\large \textsf{LHS...} \\ \\ \Longrightarrow\ \ &\csc^2A+\csc^2(90-A)\\ \\ \Longrightarrow\ \ &\left(\frac{hyp}{opp}\right)^2+\left(\frac{hyp}{adj}\right)^2\\ \\ \Longrightarrow\ \ &\frac{hyp^2}{opp^2}+\frac{hyp^2}{adj^2}\\ \\ \Longrightarrow\ \ &\frac{hyp^2(opp^2+adj^2)}{opp^2 \cdot adj^2}\end{aligned}$

Thus, according to Pythagoras' theorem  opp² + adj² = hyp²,

$\displaystyle \begin{aligned}\Longrightarrow\ \ &\frac{hyp^2 \cdot hyp^2}{opp^2 \cdot adj^2}\\ \\ \Longrightarrow\ \ &\frac{hyp^2}{opp^2} \cdot \frac{hyp^2}{adj^2}\\ \\ \Longrightarrow\ \ &\left(\frac{hyp}{opp}\right)^2 \cdot \left(\frac{hyp}{adj}\right)^2\\ \\ \Longrightarrow\ \ &\csc^2A \cdot \csc^2(90-A)\\ \\ \Longrightarrow\ \ &\large \textsf{...RHS}\end{aligned}$

Hence Proved!!!