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.
`sqrt((1 - cos A)/(1 + cos A)) = cosec A - cot A`
If 2 sin A – 1 = 0, show that: sin 3A = 3 sin A – 4 sin3 A
Prove that:
`cot^2A/(cosecA - 1) - 1 = cosecA`
Prove the following identities:
`(sin theta + 1 - cos theta)/(cos theta - 1 + sin theta) = (1 + sin theta)/(cos theta)`
Prove the following identity :
`(1 + sinθ)/(cosecθ - cotθ) - (1 - sinθ)/(cosecθ + cotθ) = 2(1 + cotθ)`
If secθ + tanθ = m , secθ - tanθ = n , prove that mn = 1
Find A if tan 2A = cot (A-24°).
1 + cot2θ = ?
`sqrt((1 - cos^2theta) sec^2 theta) = tan theta`
If sin A = `1/2`, then the value of sec A is ______.
