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प्रश्न
If x = a cos θ and y = b sin θ, then b2x2 + a2y2 =
पर्याय
a2 b2
ab
a4 b4
a2 + b2
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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
(1 + tan θ + sec θ) (1 + cot θ − cosec θ) = ______.
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
`(sec^2 theta-1) cot ^2 theta=1`
`(1+tan^2theta)(1+cot^2 theta)=1/((sin^2 theta- sin^4theta))`
Write True' or False' and justify your answer the following :
The value of sin θ+cos θ is always greater than 1 .
\[\frac{\sin \theta}{1 + \cos \theta}\]is equal to
If x = r sin θ cos ϕ, y = r sin θ sin ϕ and z = r cos θ, then
Prove the following identity :
`cosA/(1 - tanA) + sinA/(1 - cotA) = sinA + cosA`
Prove the following identity :
`(1 + cotA)^2 + (1 - cotA)^2 = 2cosec^2A`
Prove the following identity :
`(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA) = 2`
Prove the following identity :
`(cosecθ)/(tanθ + cotθ) = cosθ`
Without using trigonometric identity , show that :
`cos^2 25^circ + cos^2 65^circ = 1`
If x = r sin θ cos Φ, y = r sin θ sin Φ and z = r cos θ, prove that x2 + y2 + z2 = r2.
Without using trigonometric table, prove that
`cos^2 26° + cos 64° sin 26° + (tan 36°)/(cot 54°) = 2`
Prove that `(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`
Prove that `cot^2 "A" [(sec "A" - 1)/(1 + sin "A")] + sec^2 "A" [(sin"A" - 1)/(1 + sec"A")]` = 0
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
Prove that (1 – cos2A) . sec2B + tan2B(1 – sin2A) = sin2A + tan2B
If 1 + sin2θ = 3 sin θ cos θ, then prove that tan θ = 1 or `1/2`.
Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.
