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Question
Prove that `cos θ/sin(90° - θ) + sin θ/cos (90° - θ) = 2`.
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Solution
LHS = `cos θ/sin(90° - θ) + sin θ/cos (90° - θ) `
= `cos θ/cos θ + sin θ/sin θ`
= 1 + 1 = 2
= RHS
Hence proved.
RELATED QUESTIONS
if `cosec theta - sin theta = a^3`, `sec theta - cos theta = b^3` prove that `a^2 b^2 (a^2 + b^2) = 1`
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cosec A(1 + cos A) (cosec A – cot A) = 1
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Write the value of `3 cot^2 theta - 3 cosec^2 theta.`
Prove the following identity :
`sqrt((secq - 1)/(secq + 1)) + sqrt((secq + 1)/(secq - 1))` = 2 cosesq
Prove that cosec2 (90° - θ) + cot2 (90° - θ) = 1 + 2 tan2 θ.
Prove the following identities.
sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
Prove the following that:
`tan^3θ/(1 + tan^2θ) + cot^3θ/(1 + cot^2θ)` = secθ cosecθ – 2 sinθ cosθ
