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Trigonometry
Prove that (sin^<sup>4</sup>θ – cos^<sup>4</sup>θ +1) cosec^<sup>2</sup>θ = 2
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This discussion was modified 1 year, 9 months ago by
Lucky Chauhan. Reason: Prove that (sin^4(θ) – cos^4(θ )+1) cosec^2(θ )= 2
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This discussion was modified 1 year, 9 months ago by
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1 Reply
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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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