Question

Prove the following:

(i) sinθ sin(90° - θ) - cosθ cos (90° - θ) = 0

(ii) \(\frac{cos(90° - θ) sec(90° - θ) tanθ}{cosec(90° - θ) sin(90° - θ) cot(90° - θ)}+ \frac{tan(90° - θ)}{cotθ} = 2\)

(iii) \(\frac{tan(90° - θ) cot A}{cosec^2A} - cos^2A = 0\)

Answer 1

In the given parts use:

sin(90° - θ) = cosθ

cos(90° - θ) = sinθ

sec(90° - θ) = cosecθ

cosec(90° - θ) = secθ

tan(90° - θ) = cotθ

cot(90° - θ) = tanθ

(i) sinθ sin(90° - θ) - cosθ cos(90° - θ) = 0

solve LHS

Which is equal to RHS.

(ii) 

solve LHS, Use: sinθ = \(\frac{1}{cosecθ}\), \(secθ = \frac{1}{cosθ}\)

solve,

Which is equal to RHS.

(iii)  \(\frac{tan(90° - θ) cotA}{cosec^2A} - cos^2A =0\)

solve LHS, use:

\(tanθ =\frac{sinθ}{cosθ}\), cotθ = \(\frac{cosθ}{sinθ}\)

sinθ = \(\frac{1}{cosecθ}\),  secθ =\(\frac{1}{cosθ}\)

solve,

Which is equal  to RHS.