Question

cosec^2 thetha - cos^2 thetha/cot^2thetha = sec^2 thetha - sin^2 thetha
prove that​

Answer 1

Answer:

We have,

LHS = tan 2 θ+cot 2 θ+2

⇒ LHS = tan 2θ+cot 2θ+2tanθcotθ

⇒ LHS = (tanθ+cotθ) 2

⇒ LHS = ( cosθsinθ +sinθcosθ) 2

⇒ LHS = ( sinθcosθsin 2θ+cos 2 θ ) 2

⇒ LHS = ( sinθcosθ1 ) 2

⇒ LHS = sin 2 θcos 2 θ1

=cosec 2 θsec 2θ=RHS

ALTERNATIVELY 1

We have,

LHS = tan 2θ+cot 2θ+2=(1+tan 2 θ)+(1+cot 2θ)

⇒ LHS = sec 2 θ+cosec 2 θ= cos 2θ1 + sin 2θ1

= cos 2 θsin 2 θsin 2 θ+cos 2θ

⇒ LHS = cos 2 θsin 2 θ1

=cosec 2θsec 2 θ=RHS

ALTERNATIVELY 2

LHS = tan 2 θ+cot 2 θ+2

⇒ LHS = 1 + tan 2 θ+cot 2 θ + 1

⇒ LHS = 1 + tan 2 θ+cot 2 θ+tan 2 +θcot 2 θ

[∵tan 2 θcot 2 θ=1]

⇒ LHS = (1 + tan 2 θ) + cot 2 θ(1 + tan 2θ)

⇒ LHS = (1 + tan 2 θ)(1 + cot 2 θ) = sec 2θcosec 2 θ=RHS

ALTERNATIVELY 3

RHS = sec 2 θ+cosec 2θ

⇒ RHS = (1 + tan 2 θ)(1 + cot 2 θ)

⇒ RHS = 1 + tan

2 θ+cot 2 θ+tan 2 θcot 2 θ

⇒ RHS = 1 + tan

2 θ+cot 2co θ+1=tan 2 θ+cot 2 θ+2=LHS

Answer 2

Step-by-step explanation:

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