Question

show that tan^2teta+cot^2teta+2=sec^2teta+cosec^2teta​

Answer 1

We have,

LHS = tan

2

θ+cot

2

θ+2

⇒ LHS = tan

2

θ+cot

2

θ+2tanθcotθ

⇒ LHS = (tanθ+cotθ)

2

⇒ LHS = (

cosθ

sinθ

+

sinθ

cosθ

)

2

⇒ LHS = (

sinθcosθ

sin

2

θ+cos

2

θ

)

2

⇒ LHS = (

sinθcosθ

1

)

2

⇒ LHS =

sin

2

θcos

2

θ

1

=cosec

2

θsec

2

θ=RHS

ALTERNATIVELY 1

We have,

LHS = tan

2

θ+cot

2

θ+2=(1+tan

2

θ)+(1+cot

2

θ)

⇒ LHS = sec

2

θ+cosec

2

θ=

cos

2

θ

1

+

sin

2

θ

1

=

cos

2

θsin

2

θ

sin

2

θ+cos

2

θ

⇒ LHS =

cos

2

θsin

2

θ

1

=cosec

2

θsec

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

2

θ)

⇒ LHS = (1 + tan

2

θ)(1 + cot

2

θ) = sec

2

θcosec

2

θ=RHS

ALTERNATIVELY 3

RHS = sec

2

θ+cosec

2

θ

⇒ 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

2

θ+1=tan

2

θ+cot

2

θ+2=LHS