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Question
Prove that sin2 θ + cos4 θ = cos2 θ + sin4 θ.
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Solution
L.H.S. = sin2 θ + cos4 θ
= 1 - cos2 θ + cos4 θ
= 1 - cos2 θ (1 - cos2 θ)
= 1 - (1 - sin2 θ) sin2 θ
= 1 - sin2 θ + sin4 θ
= cos2 θ + sin4 θ
= R.H.S.
Hence proved.
RELATED QUESTIONS
Prove the following trigonometric identities:
(i) (1 – sin2θ) sec2θ = 1
(ii) cos2θ (1 + tan2θ) = 1
Prove the following identities:
(1 + tan A + sec A) (1 + cot A – cosec A) = 2
`((sin A- sin B ))/(( cos A + cos B ))+ (( cos A - cos B ))/(( sinA + sin B ))=0`
Write the value of `cosec^2 theta (1+ cos theta ) (1- cos theta).`
If `sin theta = x , " write the value of cot "theta .`
Prove that `(tan^2"A")/(tan^2 "A"-1) + (cosec^2"A")/(sec^2"A"-cosec^2"A") = (1)/(1-2 co^2 "A")`
Prove that `((1 - cos^2 θ)/cos θ)((1 - sin^2θ)/(sin θ)) = 1/(tan θ + cot θ)`
`5/(sin^2θ) - 5cot^2θ`, complete the activity given below.
Activity:
`5/(sin^2θ) - 5cot^2θ`
= `square (1/(sin^2θ) - cot^2θ)`
= `5(square - cot^2θ) ...[1/(sin^2θ) = square]`
= 5(1)
= `square`
Prove that `sec^2A - "cosec"^2A = (2sin^2A - 1)/(sin^2A *cos^2A)`.
Prove that `(cot A - cos A)/(cot A + cos A) = (cos^2 A)/(1 + sin A)^2`
