Advertisements
Advertisements
प्रश्न
If x = a cos θ and y = b sin θ, then b2x2 + a2y2 =
पर्याय
a2 b2
ab
a4 b4
a2 + b2
Advertisements
उत्तर
Given:
`x= a cosθ, y= b sin θ`
So,
`b^2 x^2+a^2 y^2`
= `b^2(a cos)^2+a^2(b sin θ)^2`
=` b^2 a^2 cos^2θ+a^2 b^2 sin^2θ`
=`b^2a^2 (cos^2 θ+sin^2θ)`
We know that,
`sin^2θ+cos^2θ=1`
Therefore,` b^2x^2+a^2y^2=a^2b^2`
APPEARS IN
संबंधित प्रश्न
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
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 - x^2/c^2 = 1`
Prove the following identities:
(1 – tan A)2 + (1 + tan A)2 = 2 sec2A
Prove the following identities:
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
`sin theta (1+ tan theta) + cos theta (1+ cot theta) = ( sectheta+ cosec theta)`
`(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))=1`
If a cos `theta + b sin theta = m and a sin theta - b cos theta = n , "prove that "( m^2 + n^2 ) = ( a^2 + b^2 )`
`If sin theta = cos( theta - 45° ),where theta " is acute, find the value of "theta` .
If \[\cos A = \frac{7}{25}\] find the value of tan A + cot A.
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
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`
Find x , if `cos(2x - 6) = cos^2 30^circ - cos^2 60^circ`
If sec θ = x + `1/(4"x"), x ≠ 0,` find (sec θ + tan θ)
If A = 30°, verify that `sin 2A = (2 tan A)/(1 + tan^2 A)`.
Prove that cot θ. tan (90° - θ) - sec (90° - θ). cosec θ + 1 = 0.
Prove that identity:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
Prove that: `1/(sec θ - tan θ) = sec θ + tan θ`.
Prove the following identities.
`(cot theta - cos theta)/(cot theta + cos theta) = ("cosec" theta - 1)/("cosec" theta + 1)`
