• Math

  Trigonometry

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

  Prove that (sin^⁴θ – cos^⁴θ +1) cosec^²θ = 2

  • This discussion was modified 1 year, 10 months ago by  Lucky Chauhan. Reason: Prove that (sin^4(θ) – cos^4(θ )+1) cosec^2(θ )= 2

  Ananya replied 1 year, 10 months ago 2 Members · 1 Reply
• 1 Reply

• 

  Ananya

  Member

  June 23, 2023 at 7:17 pm

  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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