Prove that 1/sin^2A -1/tan^2A= 1

Mathematics

1 answer:

Vika [28.1K]2 years ago

4 0

Answer:

Step-by-step explanation:

$LHS =\dfrac{1}{Sin^{2} \ A }-\dfrac{1}{Tan^{2} \ A }\\\\\\ = \dfrac{1}{sin^{2} \ A}- \dfrac{1}{\dfrac{Sin^{2} \ A}{Cos^{2} \ A}}\\\\\\= \dfrac{1}{sin^{2} \ A } - \dfrac{Cos^{2} \ A}{Sin^{2} \ A}\\\\\\= \dfrac{1-Cos^{2} \ A}{Sin^{2} \ A}\\\\\\= \dfrac{Sin^{2} \ A}{Sin^{2} \ A}\\\\\\= 1 = \ RHS$

Hint: 1 - Cos² A = Sin² A

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