Advertisements
Advertisements
Question
Prove the following identity :
`(cosecθ)/(tanθ + cotθ) = cosθ`
Advertisements
Solution
LHS = `(cosecθ)/(tanθ + cotθ)`
= `(1/sinθ)/(sinθ/cosθ + cosθ/sinθ)`
= `(1/sinθ)/((sin^2θ + cos^2θ)/(cosθsinθ))` = `(1/sinθ)/(1/(cosθsinθ)`
= `1/sinθ xx (cosθsinθ)/1 = cosθ`
APPEARS IN
RELATED QUESTIONS
Prove that `(sin theta)/(1-cottheta) + (cos theta)/(1 - tan theta) = cos theta + sin theta`
Prove the following identities:
(cos A + sin A)2 + (cos A – sin A)2 = 2
Prove the following identities:
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
Find the value of `(cos 38° cosec 52°)/(tan 18° tan 35° tan 60° tan 72° tan 55°)`
If a cos θ + b sin θ = 4 and a sin θ − b sin θ = 3, then a2 + b2 =
Without using trigonometric table , evaluate :
`cos90^circ + sin30^circ tan45^circ cos^2 45^circ`
If cosθ = `5/13`, then find sinθ.
Proved that cosec2(90° - θ) - tan2 θ = cos2(90° - θ) + cos2 θ.
Prove the following identities.
`(cot theta - cos theta)/(cot theta + cos theta) = ("cosec" theta - 1)/("cosec" theta + 1)`
Statement 1: sin2θ + cos2θ = 1
Statement 2: cosec2θ + cot2θ = 1
Which of the following is valid?
