Advertisements
Advertisements
Question
Prove the following identity :
`(tanθ + sinθ)/(tanθ - sinθ) = (secθ + 1)/(secθ - 1)`
Advertisements
Solution
LHS = `(tanθ + sinθ)/(tanθ - sinθ)`
= `(sinθ/cosθ + sinθ)/(sinθ/cosθ - sinθ) = (sinθ + sinθcosθ)/(sinθ + sinθcosθ)`
= `(sinθ(1 + cosθ))/sin(1 + cosθ) = (1 + cosθ)/(1 - cosθ)`
= `(1 + 1/secθ)/(1 - 1/secθ) = ((secθ + 1)/secθ)/((secθ - 1)/secθ)`
= `(secθ + 1)/(secθ - 1)`
APPEARS IN
RELATED QUESTIONS
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
`(sintheta - 2sin^3theta)/(2costheta - costheta) =tan theta`
Prove that (1 + cot θ – cosec θ)(1+ tan θ + sec θ) = 2
Prove the following identities:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
Find the value of sin ` 48° sec 42° + cos 48° cosec 42°`
Prove the following identity :
`1/(cosA + sinA - 1) + 2/(cosA + sinA + 1) = cosecA + secA`
Prove the following identity :
`(sec^2θ - sin^2θ)/tan^2θ = cosec^2θ - cos^2θ`
Without using trigonometric table , evaluate :
`cos90^circ + sin30^circ tan45^circ cos^2 45^circ`
Prove that: 2(sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = 0.
Without using the trigonometric table, prove that
tan 10° tan 15° tan 75° tan 80° = 1
If 5x = sec θ and `5/x` = tan θ, then `x^2 - 1/x^2` is equal to
