Question

prove that :- tan²theta -cot²theta = (tan theta -cot theta )/(sin theta.cos theta)​
its urgent

Answer 1

To prove :-

tan²θ - cot²θ = ( tan θ - cot θ ) / ( sin θ . cos θ )

Proof :-

Taking LHS

→ LHS =  tan²θ - cot²θ

putting tan θ = sin θ / cos θ  and cot θ = cos θ / sin θ

→ (sin²θ / cos²θ) - (cos²θ / sin²θ)

Taking LCM

→ ( (sin²θ)² - (cos²θ)² ) / ( sin²θ . cos²θ )

using algebraic identity a² - b² = ( a + b ) ( a - b ) in Numerator

→ (( sin²θ + cos²θ ) ( sin²θ - cos²θ )) / ( sin²θ . cos²θ )

using trigonometric identity

sin²θ + cos²θ = 1

→ ( 1 ( sin²θ - cos²θ )) / ( sin²θ . cos²θ )

→ ( sin²θ - cos²θ ) / ( sin²θ . cos²θ )

Now,

Taking RHS

→ RHS =  ( tan θ - cot θ ) / ( sin θ . cos θ )

putting tan θ = sin θ / cos θ and cot θ = cos θ / sin θ

→ ( (sin θ / cos θ) - (cos θ / sin θ) ) / ( sin θ . cos θ )

→ ( ( sin²θ - cos²θ ) / ( sin θ . cos θ ) ) / ( sin θ . cos θ )

→ ( sin²θ - cos²θ ) / ( sin²θ . cos²θ )

so,

LHS = RHS

Proved .