Advertisements
Advertisements
Question
Prove that `sqrt((1 + sin A)/(1 - sin A))` = sec A + tan A.
Advertisements
Solution
LHS = `sqrt((1 + sin A)/(1 - sin A))`
= `sqrt((1 + sin A)/(1 - sin A) xx (1 + sin A)/(1 + sin A)`
= `sqrt((1 + sin A)^2/(1 - sin^2 A))`
= `sqrt((1 + sin A)^2/cos^2 A)`
= `(1 + sin A)/cos A`
= sec A + tan A = RHS
RELATED QUESTIONS
If `sec alpha=2/sqrt3` , then find the value of `(1-cosecalpha)/(1+cosecalpha)` where α is in IV quadrant.
If cosθ + sinθ = √2 cosθ, show that cosθ – sinθ = √2 sinθ.
Prove the following trigonometric identities.
`sin theta/(1 - cos theta) = cosec theta + cot theta`
Prove the following trigonometric identities.
`(sec A - tan A)/(sec A + tan A) = (cos^2 A)/(1 + sin A)^2`
Prove the following trigonometric identities
If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2 − y2 = a2 − b2
Prove the following identities:
(1 – tan A)2 + (1 + tan A)2 = 2 sec2A
Prove the following identities:
`cot^2A/(cosecA + 1)^2 = (1 - sinA)/(1 + sinA)`
If x = r cos A cos B, y = r cos A sin B and z = r sin A, show that : x2 + y2 + z2 = r2
`(1 + cot^2 theta ) sin^2 theta =1`
`1/((1+ sin θ)) + 1/((1 - sin θ)) = 2 sec^2 θ`
`{1/((sec^2 theta- cos^2 theta))+ 1/((cosec^2 theta - sin^2 theta))} ( sin^2 theta cos^2 theta) = (1- sin^2 theta cos ^2 theta)/(2+ sin^2 theta cos^2 theta)`
Write the value of `3 cot^2 theta - 3 cosec^2 theta.`
What is the value of (1 + cot2 θ) sin2 θ?
If cosec θ − cot θ = α, write the value of cosec θ + cot α.
cos4 A − sin4 A is equal to ______.
Prove the following identity :
`(secA - 1)/(secA + 1) = sin^2A/(1 + cosA)^2`
Without using trigonometric identity , show that :
`sec70^circ sin20^circ - cos20^circ cosec70^circ = 0`
Prove that `sqrt(2 + tan^2 θ + cot^2 θ) = tan θ + cot θ`.
Prove that: `(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(sin^2 A - cos^2 A)`.
sin4A – cos4A = 1 – 2cos2A. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` ...`[sin^2"A" + square = 1]`
= `square` – cos2A ...[sin2A = 1 – cos2A]
= `square`
= R.H.S.
