Question

prove that tan^2 a ×(coseca-1/1+cosa)+cosec^2 a (cosa-1/1+coseca =0

Answer 1

GIVEN :

Prove that $tan^2 a\times (\frac{coseca-1}{1+cosa})+cosec^2 a\times (\frac{cosa-1}{1+coseca}) =0$

TO PROVE :

The given equation $tan^2 a\times (\frac{coseca-1}{1+cosa})+cosec^2a\times (\frac{cosa-1}{1+coseca}) =0$ is true

SOLUTION :

Given equation is $tan^2 a\times (\frac{coseca-1}{1+cosa})+cosec^2 a \times (\frac{cosa-1}{1+coseca}) =0$

We have to prove that the given equation is true.

Now taking the LHS

$tan^2 a\times (\frac{coseca-1}{1+cosa})+cosec^2a\times (\frac{cosa-1}{1+coseca} )$

Multiplying and dividing by its denominators conjugate for both the terms,

$=\frac{sin^2a}{cos^2a}\times (\frac{coseca-1}{1+cosa})\times (\frac{1-cosa}{1-cosa})+cosec^2 a\times (\frac{cosa-1}{1+coseca})\times (\frac{1-coseca}{1-coseca})$

By using the algebraic identity :

$(a+b)(a-b)=a^2-b^2$

$=\frac{sin^2a}{cos^2a}\times (\frac{(coseca-1)(1-cosa)}{1^2-cos^2a})+cosec^2 a (\frac{(cosa-1)(1-coseca)}{1^2-cosec^2a})$

By using the trignometric identities

• $tanx=\frac{sinx}{cosx}$

• $cotx=\frac{cosx}{sinx}$

• $1-cos^2x=sin^2x$

• $1-cosec^2x=-cot^2x$
• $cosec^2x=\frac{1}{sin^2x}$

$=\frac{1}{cos^2a}\times (coseca-cota-1+cosa)+\frac{1}{sin^2a}(\frac{cosa-cota-1+coseca}{-cot^2a})$

$=\frac{1}{cos^2a}\times (coseca-cota-1+cosa)+\frac{1}{sin^2a}(\frac{cosa-cota-1+coseca}{-\frac{cos^2a}{sin^2a}})$

$=\frac{1}{cos^2a}\times (coseca-cota-1+cosa)+\frac{cosa-cota-1+coseca}{-cos^2a}$

$=\frac{1}{cos^2a}\times (coseca-cota-1+cosa)+(\frac{-cosa+cota+1-coseca}{cos^2a})$

$=\frac{coseca-cota-1+cosa-cosa+cota+1-coseca}{cos^2a}$

$=\frac{0}{cos^2a}$

= 0 = RHS

LHS = RHS

∴ $tan^2 a\times (\frac{coseca-1}{1+cosa})+cosec^2 a\times (\frac{cosa-1}{1+coseca}) =0$