Question

(sec^2x+tan^2x)(cose^2x+ cot^2x)=1+2sec^2xcosec^2x

Answer 1

Answer:

sec²A = 1 + tan²A and cosec²A = 1 + cot²A

In the question -

→ ( sec²x + tan²x )( cosec²x + cot²x )

→ ( 1 + tan²x + tan²x )( 1 + cot²x + cot²x )

→ ( 1 + 2tan²x )( 1 + 2cot²x )

→ 1 + 2cot²x + 2tan²x + 4tan²x.cot²x

→ 1 + 2[ cot²x + tan²x + 2tan²x.cot²x ]

→ 1 + 2( cotx + tanx )²

→ 1 + 2[ cosx/sinx + sinx/cosx ]²

→ 1 + 2( sin²x + cos²x )²/( sinx.cosx )²

→ 1 + 2( 1/sin²x.cos²x )

→ 1 + 2( sec²x.cosec²x )

→ 1 + 2sec²x cosec²x

Hence proved