Advertisements
Advertisements
Question
If x = a sec θ and y = b tan θ, then b2x2 − a2y2 =
Options
ab
a2 − b2
a2 + b2
a2 b2
Solution
Given:
`x= a secθ, y=b tanθ`
So,
`b^2x^2-a^2 y^2`
=` b^2(a secθ)^2-a^2(btan θ)^2`
= `b^2 a^2 sec^2 θ-a^2 b^2 tan^2θ`
=` b^2 a^2 (sec^2θ-tan^2 θ)`
We know that,`
`sec^2θ-tan^2θ=1`
Therfore,
`b^2x^2-a^2y^2=a^2b^2`
APPEARS IN
RELATED QUESTIONS
If secθ + tanθ = p, show that `(p^{2}-1)/(p^{2}+1)=\sin \theta`
Prove that `cosA/(1+sinA) + tan A = secA`
Prove the following trigonometric identities
`cos theta/(1 - sin theta) = (1 + sin theta)/cos theta`
Prove the following trigonometric identities.
`(1 - tan^2 A)/(cot^2 A -1) = tan^2 A`
Prove the following trigonometric identities.
`((1 + sin theta - cos theta)/(1 + sin theta + cos theta))^2 = (1 - cos theta)/(1 + cos theta)`
Prove the following identities:
`(1 + sinA)/cosA + cosA/(1 + sinA) = 2secA`
Prove that:
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
If sin A + cos A = m and sec A + cosec A = n, show that : n (m2 – 1) = 2 m
\[\frac{\sin \theta}{1 + \cos \theta}\]is equal to
If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, then\[\frac{x^2}{a^2} + \frac{y^2}{b^2}\]
Simplify
sin A `[[sinA -cosA],["cos A" " sinA"]] + cos A[[ cos A" sin A " ],[-sin A" cos A"]]`
Prove the following identity :
`(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA) = 2`
Prove that tan2Φ + cot2Φ + 2 = sec2Φ.cosec2Φ.
Prove that `sqrt(2 + tan^2 θ + cot^2 θ) = tan θ + cot θ`.
If sin θ + cos θ = `sqrt(3)`, then prove that tan θ + cot θ = 1
Prove that sin4A – cos4A = 1 – 2cos2A
Prove that `"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")` = 1 + tan A + cot A = sec A . cosec A + 1
`sqrt((1 - cos^2theta) sec^2 theta) = tan theta`
Prove the following:
`sintheta/(1 + cos theta) + (1 + cos theta)/sintheta` = 2cosecθ
