Advertisements
Advertisements
Question
If tan A + sin A = m and tan A − sin A = n, then show that `m^2 - n^2 = 4 sqrt (mn)`.
Advertisements
Solution
Here,
m2 − n2 = (tan A + sin A)2 − (tan A − sin A)2
m2 − n2 = (tan A + sin A + tan A − sin A) (tan A + sin A − tan A + sin A)
m2 − n2 = (2 tan A)(2 sin A)
m2 − n2 = 4 tan A sin A ...(1)
Also,
`4 sqrt (mn) = 4 sqrt ((tan A + sin A)( tan A - sin A))`
= `4 sqrt(tan^2 A - sin^2 A)`
= `4 sqrt((sin^2 A)/(cos^2 A) - sin^2 A)`
= `4 sin A sqrt((1 - cos^2 A)/(cos^2 A))`
= `4 sin A sqrt((sin^2 A)/(cos^2 A))`
= `4 sin A * (sin A)/(cos A)`
= `4 sin A . tan A` ...(2)
Using equation (1) and equation (2) we get the required conditions.
Hence proved.
RELATED QUESTIONS
If m=(acosθ + bsinθ) and n=(asinθ – bcosθ) prove that m2+n2=a2+b2
Prove the following trigonometric identities.
sec A (1 − sin A) (sec A + tan A) = 1
`(sectheta- tan theta)/(sec theta + tan theta) = ( cos ^2 theta)/( (1+ sin theta)^2)`
`(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos ^3 θ - sin^3 θ)/(cos θ - sin θ) = 2`
If `m = (cos θ - sin θ)` and `n = (cos θ + sin θ)`, show that `sqrt(m/n) + sqrt(n/m) = 2/sqrt(1 - tan^2θ)`.
Prove the following identity :
tanA+cotA=secAcosecA
Without using trigonometric identity , show that :
`sec70^circ sin20^circ - cos20^circ cosec70^circ = 0`
Prove that sin2 θ + cos4 θ = cos2 θ + sin4 θ.
If `sec θ + tan θ = sqrt(3)`, complete the activity to find the value of sec θ – tan θ.
Activity:
`square = 1 + tan^2θ` ...[Fundamental trigonometric identity]
`square - tan^2θ = 1`
`(sec θ + tan θ) . (sec θ - tan θ) = square`
`sqrt(3) . (sec θ - tan θ) = 1`
`(sec θ - tan θ) = square`
If tan α + cot α = 2, then tan20α + cot20α = ______.
