Question

prove that cot square A cosec square B -cot square B cosec square A = cot square A- cot square B​

Answer 1

To prove--->

Cot²A Cosec²B - Cot²B Cosec²A = Cot²A - Cot²B

Proof--->

LHS

= Cot²A Cosec²B - Cot²B Cosec²A

We have an identity of trigonometery,

1 + Cot²θ = Cosec²θ , applying it here we get,

= Cot²A ( 1 + Cot²B ) - Cot²B ( 1 + Cot²A )

= Cot²A + Cot²A Cot²B - ( Cot²B + Cot²A Cot²B )

= Cot²A + Cot²A Cot²B - Cot²B - Cot²A Cot²B

= Cot²A - Cot²B = RHS

Additional information--->

1) Sin²θ + Cos²θ = 1

2) 1 + tan²θ = Sec²θ

3) 1 + Cot²θ = Cosec²θ

4) Sin( 90° - θ ) = Cosθ

5) Cos ( 90° - θ ) = Sinθ

6) tan ( 90° - θ ) = Cotθ

7) Cot ( 90° - θ ) = tanθ

8) Sec ( 90° - θ ) = Cosecθ

9) Cosec ( 90° - θ ) = Secθ