Advertisements
Advertisements
Question
`1+((tan^2 theta) cot theta)/(cosec^2 theta) = tan theta`
Advertisements
Solution
LHS = `((1+ tan^2 theta) cot theta)/ (cosec^2 theta) `
= ` (sec^2 theta cot theta)/(cosec^2 theta )`
= `(1/cos^2thetaxxcos theta/sin theta)/(1/sin^2 theta)`
= `1/(cos theta sin theta) xx sin^2 theta`
= `sintheta/costheta`
= ` tan theta`
= RHS
Hence, LHS = RHS
APPEARS IN
RELATED QUESTIONS
Prove the following trigonometric identities.
sec A (1 − sin A) (sec A + tan A) = 1
Prove the following trigonometric identities.
`(cot A + tan B)/(cot B + tan A) = cot A tan B`
Prove the following identities:
`(1 + sin A)/(1 - sin A) = (cosec A + 1)/(cosec A - 1)`
Prove the following identities:
`1/(tan A + cot A) = cos A sin A`
Prove the following identities:
`cosA/(1 + sinA) + tanA = secA`
If tan A = n tan B and sin A = m sin B, prove that:
`cos^2A = (m^2 - 1)/(n^2 - 1)`
`(1-cos^2theta) sec^2 theta = tan^2 theta`
`sqrt((1+sin theta)/(1-sin theta)) = (sec theta + tan theta)`
If a cos `theta + b sin theta = m and a sin theta - b cos theta = n , "prove that "( m^2 + n^2 ) = ( a^2 + b^2 )`
Write the value of cosec2 (90° − θ) − tan2 θ.
Prove the following identity :
`sec^4A - sec^2A = sin^2A/cos^4A`
Prove that `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec(90^circ - A) cosec(90^circ - A)`
If cosθ = `5/13`, then find sinθ.
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
Prove that `(cot "A" + "cosec A" - 1)/(cot "A" - "cosec A" + 1) = (1 + cos "A")/sin "A"`
Prove that `sin^2 θ/ cos^2 θ + cos^2 θ/sin^2 θ = 1/(sin^2 θ. cos^2 θ) - 2`.
Prove that : `tan"A"/(1 - cot"A") + cot"A"/(1 - tan"A") = sec"A".cosec"A" + 1`.
Prove the following identities.
(sin θ + sec θ)2 + (cos θ + cosec θ)2 = 1 + (sec θ + cosec θ)2
Prove that `(cos^2theta)/(sintheta) + sintheta` = cosec θ
Prove that
sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A
