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Question
Prove that `(sin θ tan θ)/(1 - cos θ) = 1 + sec θ.`
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Solution
LHS = `( sin θ tan θ)/(1 - cos θ)`
= `(sin θ. (sin θ)/(cos θ))/(1 - cos θ)`
= `sin^2 θ/(cos θ( 1 - cos θ))`
= `((1 - cos θ)(1 + cos θ))/(cos θ(1 - cos θ))`
= `(1 + cos θ)/(cos θ) = 1/(cos θ) + cos θ/cos θ`
= sec θ + 1
= RHS
Hence proved.
RELATED QUESTIONS
Prove that ` \frac{\sin \theta -\cos \theta +1}{\sin\theta +\cos \theta -1}=\frac{1}{\sec \theta -\tan \theta }` using the identity sec2 θ = 1 + tan2 θ.
If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, show that `x^2/a^2 + y^2/b^2 - z^2/c^2 = 1`
Prove the following identities:
`1/(1 - sinA) + 1/(1 + sinA) = 2sec^2A`
Prove the following identities:
`1/(sinA + cosA) + 1/(sinA - cosA) = (2sinA)/(1 - 2cos^2A)`
`(sin theta +cos theta )/(sin theta - cos theta)+(sin theta- cos theta)/(sin theta + cos theta) = 2/((sin^2 theta - cos ^2 theta)) = 2/((2 sin^2 theta -1))`
If sec θ + tan θ = x, write the value of sec θ − tan θ in terms of x.
Prove the following identity:
tan2A − sin2A = tan2A · sin2A
tan θ cosec2 θ – tan θ is equal to
If sin A = `1/2`, then the value of sec A is ______.
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
