Question

(sinθ+secθ)² + (cosθ + cosecθ)² = (1+secθ cosecθ)². ​

Answer 1

L.H.S.
=(sinθ+secθ)²+(cosθ+cosecθ)²
=sin²θ+2sinθsecθ+sec²θ+cos²θ+2cosθcosecθ+cosec²θ
=sin²θ+cos²θ+2(sinθsecθ+cosθcosecθ)+sec²θ+cosec²θ
=1+2(sinθ/cosθ+cosθ/sinθ)+(1/cos²θ+1/sin²θ)
=1+2(sin²θ+cos²θ)/sinθcosθ+(sin²θ+cos²θ)/sin²θcos²θ
=1+2/sinθcosθ+1/sin²θcos²θ
=1+2secθcosecθ+sec²θcosec²θ
R.H.S.
=(1+secθcosecθ)²
=1²+2×1×(secθcosecθ)+(secθcosecθ)²
=1+2secθcosecθ+sec²θcosec²θ
∴, L.H.S.=R.H.S. (Proved)

Answer 2

Answer:

Step-by-step explanation:

L.H.S.

=(sinθ+secθ)²+(cosθ+cosecθ)²

=sin²θ+2sinθsecθ+sec²θ+cos²θ+2cosθcosecθ+cosec²θ

=sin²θ+cos²θ+2(sinθsecθ+cosθcosecθ)+sec²θ+cosec²θ

=1+2(sinθ/cosθ+cosθ/sinθ)+(1/cos²θ+1/sin²θ)

=1+2(sin²θ+cos²θ)/sinθcosθ+(sin²θ+cos²θ)/sin²θcos²θ

=1+2/sinθcosθ+1/sin²θcos²θ

=1+2secθcosecθ+sec²θcosec²θ    -----(1)

R.H.S.

=(1+secθcosecθ)²

=1²+2×1×(secθcosecθ)+(secθcosecθ)²

=1+2secθcosecθ+sec²θcosec²θ    ------(2)

from 1 and 2

LHS=RHS