Advertisements
Advertisements
Question
Prove the following identities: cot θ - tan θ = `(2 cos^2 θ - 1)/(sin θ cos θ)`.
Advertisements
Solution
LHS = cot θ - tan θ
= `cos θ/sin θ - sin θ/cos θ`
= `(cos^2 θ - sin^2 θ)/(sin θ. cos θ)`
= `(cos^2 θ - (1 - cos^2 θ))/(sin θ. cos θ)`
= `(2cos^2 θ - 1)/(sin θ. cos θ)`
= RHS
Hence proved.
RELATED QUESTIONS
Prove the following trigonometric identities.
tan2 θ − sin2 θ = tan2 θ sin2 θ
Prove the following trigonometric identities.
`(1 + cos A)/sin^2 A = 1/(1 - cos A)`
Prove the following identities:
`secA/(secA + 1) + secA/(secA - 1) = 2cosec^2A`
Prove that:
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
Prove that:
`"tan A"/(1 + "tan"^2 "A")^2 + "Cot A"/(1 + "Cot"^2 "A")^2 = "sin A cos A"`.
Prove the following identity :
`sqrt((secq - 1)/(secq + 1)) + sqrt((secq + 1)/(secq - 1))` = 2 cosesq
Without using trigonometric identity , show that :
`sec70^circ sin20^circ - cos20^circ cosec70^circ = 0`
Proved that cosec2(90° - θ) - tan2 θ = cos2(90° - θ) + cos2 θ.
If `sqrt(3)` sin θ – cos θ = θ, then show that tan 3θ = `(3tan theta - tan^3 theta)/(1 - 3 tan^2 theta)`
a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p2 – q2 is equal to
