Advertisements
Advertisements
Question
If tanA + sinA = m and tanA - sinA = n , prove that (`m^2 - n^2)^2` = 16mn
Advertisements
Solution
Consider `(m^2 - n^2) = (tanA + sinA)^2 - (tanA - sinA)^2`
⇒ [tanA + sinA - (tanA - sinA)] [tanA + sinA + (tanA - sinA)]
⇒ [2sinA][2tanA] = 4sinAtanA
Now LHS = `(m^2 - n^2)^2 = (4sinAtanA)^2 = 16sin^2Atan^2A`
Also , RHS = 16mn = 16(tanA + sinA)(tanA - sinA)
⇒ RHS = 16mn = `16(tan^2A - sin^2A) = 16(sin^2A/cos^2A - sin^2A)`
⇒ `16sin^2A((1 - cos^2A)/cos^2A) = 16sin^2A(sin^2A/cos^2A) = 16sin^2Atan^2A`
Thus , `(m^2 - n^2)^2` = 16mn
APPEARS IN
RELATED QUESTIONS
Prove the following trigonometric identities.
(sec2 θ − 1) (cosec2 θ − 1) = 1
Prove the following trigonometric identities.
`[tan θ + 1/cos θ]^2 + [tan θ - 1/cos θ]^2 = 2((1 + sin^2 θ)/(1 - sin^2 θ))`
Prove that:
(1 + tan A . tan B)2 + (tan A – tan B)2 = sec2 A sec2 B
Show that : `sinAcosA - (sinAcos(90^circ - A)cosA)/sec(90^circ - A) - (cosAsin(90^circ - A)sinA)/(cosec(90^circ - A)) = 0`
If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, then\[\frac{x^2}{a^2} + \frac{y^2}{b^2}\]
Prove the following identity :
`(cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)`
If tan θ = 2, where θ is an acute angle, find the value of cos θ.
Prove that sin2 θ + cos4 θ = cos2 θ + sin4 θ.
Prove that `(sin θ + "cosec" θ)/(sin θ) = 2 + cot^2θ`.
Factorize: sin3θ + cos3θ
Hence, prove the following identity:
`(sin^3θ + cos^3θ)/(sin θ + cos θ) + sin θ cos θ = 1`
