Advertisements
Advertisements
प्रश्न
Prove the following identity :
`(sec^2θ - sin^2θ)/tan^2θ = cosec^2θ - cos^2θ`
Advertisements
उत्तर
LHS = `(sec^2θ - sin^2θ)/tan^2θ`
= `(1/cos^2θ - sin^2θ)/(sin^2θ/cos^2θ)`
= `(1 - sin^2θcos^2θ)/((cos^2θ)/(sin^2θ/cos^2θ))`
= `(1 - sin^2θcos^2θ)/sin^2θ`
= `1/sin^2θ - (sin^2θcos^2θ)/(sin^2θ)`
= `cosec^2θ - cos^2θ`
APPEARS IN
संबंधित प्रश्न
Evaluate
`(sin ^2 63^@ + sin^2 27^@)/(cos^2 17^@+cos^2 73^@)`
Prove the following trigonometric identities.
`cos A/(1 - tan A) + sin A/(1 - cot A) = sin A + cos A`
Show that none of the following is an identity:
`sin^2 theta + sin theta =2`
If `sin theta = x , " write the value of cot "theta .`
Prove the following identity :
`(1 + tan^2A) + (1 + 1/tan^2A) = 1/(sin^2A - sin^4A)`
If x sin3θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ , then show that x2 + y2 = 1.
Prove that sin2 5° + sin2 10° .......... + sin2 85° + sin2 90° = `9 1/2`.
Prove that `(cos(90 - "A"))/(sin "A") = (sin(90 - "A"))/(cos "A")`
(sec θ + tan θ) . (sec θ – tan θ) = ?
Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.
