Given A is an acute angle and cosec A = √2

Given A is an acute angle and cosec A = √2, find the value of

(2 sin² A + 3 cot² A)/ (tan² A – cos² A).

Answer :

Let triangle ABC be a right angled at B and A is a acute angle.

Given that cosec A = √2

AC/BC = √2/1

Let AC = √2x

Then BC = x

In right angled triangle ABC

AC² = AB² + BC²

⇒ (√2x)² = AB² + x²

⇒ AB² = 2x² – x²

⇒ AB = x

sin A = perpendicular/ hypotenuse

= BC/AC

= 1/ √2

cot A = base/ perpendicular

= x/x

= 1

Tan A = perpendicular/ base

= BC/AB

= x/x

= 1

cos A = base/ hypotenuse

= AB/AC

= x/ √2x

= 1/√2

2 sin²A + 3 cot²A/(tan²A – cos²A) = 8

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