Question

Cos (90-theta)/1+sin(90-theta)+1+sin(90-theta)/cos(90-theta)=2csc theta

Answer 1

$\boxed{\textbf{\large{Step-by-step explanation:}}}$

To prove :

(Cos (90-Φ)/[1+sin(90-Φ)])+([1+sin(90-Φ)]/cos(90-Φ))=2cosecΦ

proof :

here first remember that ,

cos ( 90 - Φ ) = sin Φ

sin ( 90 - Φ ) = cos Φ

Therefor,

LHS = (Cos (90-Φ)/[1+sin(90-Φ)])+([1+sin(90-Φ)]/cos(90-Φ))

= (sin Φ/(1+cosΦ))+(1+cosΦ)/sinΦ)

=(sin^2Φ + ( 1+cosΦ)^2)/((1+cosΦ)sinΦ)

=(sin^2Φ+1+2cosΦ+cos^2Φ)

/((1 + cosΦ) sinΦ)

=(( sin^2Φ + cos^2Φ) + 1 + 2cosΦ)

/ ((1 + cosΦ ) sinΦ )

=( 1 + 1 + 2cosΦ ) / ((1+cosΦ)sinΦ)

=( 2 + 2 cosΦ ) / ((1+cosΦ)sinΦ)

=((2(1+cosΦ))/((1+cosΦ)sinΦ)

= 2 / sinΦ

=2cosecΦ = RHS

hence proved