Advertisements
Advertisements
प्रश्न
If x = a sin θ and y = b cos θ, what is the value of b2x2 + a2y2?
Advertisements
उत्तर
Given:
x = a sinθ and y = b cosθ
So, \[b^2 x^2 + a^2 y^2 = b^2 \left( asin\theta \right)^2 + a^2 \left( bcos\theta \right)^2 \]
\[ = a^2 b^2 \sin^2 \theta + a^2 b^2 \cos^2 \theta\]
\[ = a^2 b^2 \left( \sin^2 \theta + \cos^2 \theta \right)\]
We know that, `sin^2 θ+cos^2θ=1`
Therefore,
\[b^2 x^2 + a^2 y^2 = a^2 b^2\]
APPEARS IN
संबंधित प्रश्न
If (secA + tanA)(secB + tanB)(secC + tanC) = (secA – tanA)(secB – tanB)(secC – tanC) prove that each of the side is equal to ±1. We have,
Prove that `cosA/(1+sinA) + tan A = secA`
Prove the following identities:
sec4 A (1 – sin4 A) – 2 tan2 A = 1
` (sin theta + cos theta )/(sin theta - cos theta ) + ( sin theta - cos theta )/( sin theta + cos theta) = 2/ ((1- 2 cos^2 theta))`
`(cos theta cosec theta - sin theta sec theta )/(costheta + sin theta) = cosec theta - sec theta`
If cosec θ − cot θ = α, write the value of cosec θ + cot α.
If 5x = sec θ and \[\frac{5}{x} = \tan \theta\]find the value of \[5\left( x^2 - \frac{1}{x^2} \right)\]
The value of \[\sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}}\]
\[\frac{\sin \theta}{1 + \cos \theta}\]is equal to
Prove the following identity :
`(1 + tan^2A) + (1 + 1/tan^2A) = 1/(sin^2A - sin^4A)`
Prove the following identity :
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
Prove the following identity :
`sin^8θ - cos^8θ = (sin^2θ - cos^2θ)(1 - 2sin^2θcos^2θ)`
Without using a trigonometric table, prove that
`(cos 70°)/(sin 20°) + (cos 59°)/(sin 31°) - 8sin^2 30° = 0`.
tan θ cosec2 θ – tan θ is equal to
If (sin α + cosec α)2 + (cos α + sec α)2 = k + tan2α + cot2α, then the value of k is equal to
Prove that cos2θ . (1 + tan2θ) = 1. Complete the activity given below.
Activity:
L.H.S. = `square`
= `cos^2θ xx square` ...`[1 + tan^2θ = square]`
= `(cos θ xx square)^2`
= 12
= 1
= R.H.S.
Prove that `sec^2A - "cosec"^2A = (2sin^2A - 1)/(sin^2A *cos^2A)`.
Prove that sin6A + cos6A = 1 – 3sin2A . cos2A.
If 5 tan β = 4, then `(5 sin β - 2 cos β)/(5 sin β + 2 cos β)` = ______.
