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Activity Discussion Math Trigonometry

• # Trigonometry

Posted by on June 23, 2023 at 4:46 pm

Prove that (sin^<sup>4</sup>θ – cos^<sup>4</sup>θ +1) cosec^<sup>2</sup>θ = 2

• This discussion was modified 1 year, 1 month ago by  Lucky Chauhan. Reason: Prove that (sin^4(θ) – cos^4(θ )+1) cosec^2(θ )= 2
replied 1 year, 1 month ago 2 Members · 1 Reply
• ### Ananya

Member
June 23, 2023 at 7:17 pm
3

To prove :

(Sin^4(θ) – cos^4(θ) + 1)(cosec^2(θ) =2

Left hand side :

(Sin^4(θ) – cos^4(θ)+1)(cosec^2(θ))

=> We can write sin^4(θ) as (sin^2(θ))^2

and cos^4(θ) as (cos^2(θ))^2 .

=> ((Sin^2(θ))^2 -(cos^2(θ))^2 +1)(cosec^2(θ))

=> Now we will use the following identity: (a^2 – b^2) = (a+b)(a-b) in our equation

=> ((Sin^2(θ)-cos^2(θ))(sin^2(θ) + cos^2(θ) ) + 1)(cosec^2(θ))

=> We know that :

(sin^2(θ) + cos^2(θ))= 1 ,

Therefore,

((Sin^2(θ)-cos^2(θ)).1 +1)(cosec^2(θ))

=> We can write it as ↓

=> (Sin^2(θ) +1- cos^2(θ))(cosec^2(θ))

=> We can write sin^2(θ) as

(1- cos^2(θ)) as we know sin^2(θ)+cos^2(θ) = 1

sin^2(θ) = 1- cos^2(θ) —{eq a}

=> (sin^2(θ) + sin^2(θ))(cosec^2(θ)). {using eq a}

=> (2sin^2(θ))(cosec^2(θ))

=> Using the relation between cosec(θ) and sin(θ)

We know sin(θ)=1/(cosec(θ) or

Cosec(θ)=1/(sin(θ)

Squaring both side

cosec^2(θ)=1/(sin^2(θ)). —eq 1

=> (2sin^2(θ))/(sin^2(θ)). {using eq 1}

=> 2

Hence proved

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