Advertisements
Advertisements
Question
Eliminate θ if x = r cosθ and y = r sinθ.
Solution
x = r cosθ and y = r sinθ
Squaring on both terms,
x2 = r2cos2θ ...(1)
y2 = r2sin2θ ...(2)
Add (1) + (2).
x2 + y2 = r2sin2θ + r2cos2θ
x2 + y2 = r2(sin2θ + cos2θ)
But we know, (sin2θ + cos2θ) = 1
∴ x2 + y2 = r2
APPEARS IN
RELATED QUESTIONS
If m=(acosθ + bsinθ) and n=(asinθ – bcosθ) prove that m2+n2=a2+b2
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(cosec θ – cot θ)^2 = (1-cos theta)/(1 + cos theta)`
Prove the following trigonometric identities
`((1 + sin theta)^2 + (1 + sin theta)^2)/(2cos^2 theta) = (1 + sin^2 theta)/(1 - sin^2 theta)`
Prove the following identities:
`(1 + sinA)/cosA + cosA/(1 + sinA) = 2secA`
Prove that:
`tanA/(1 - cotA) + cotA/(1 - tanA) = secA cosecA + 1`
If sin A + cos A = m and sec A + cosec A = n, show that : n (m2 – 1) = 2 m
`(sec^2 theta-1) cot ^2 theta=1`
`1+ (cot^2 theta)/((1+ cosec theta))= cosec theta`
`1+((tan^2 theta) cot theta)/(cosec^2 theta) = tan theta`
Write the value of `(sin^2 theta 1/(1+tan^2 theta))`.
Simplify : 2 sin30 + 3 tan45.
If cos (\[\alpha + \beta\]= 0 , then sin \[\left( \alpha - \beta \right)\] can be reduced to
Prove the following identity :
`sqrt((1 + sinq)/(1 - sinq)) + sqrt((1- sinq)/(1 + sinq))` = 2secq
Prove the following identity :
`(1 + tan^2A) + (1 + 1/tan^2A) = 1/(sin^2A - sin^4A)`
Prove the following identity :
`(1 + tan^2θ)sinθcosθ = tanθ`
Prove the following identity :
`tan^2θ/(tan^2θ - 1) + (cosec^2θ)/(sec^2θ - cosec^2θ) = 1/(sin^2θ - cos^2θ)`
Evaluate:
sin2 34° + sin2 56° + 2 tan 18° tan 72° – cot2 30°
Without using the trigonometric table, prove that
tan 10° tan 15° tan 75° tan 80° = 1
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
