Advertisements
Advertisements
Question
If tan A + sin A = m and tan A - sin A = n, then show that m2 - n2 = 4 `sqrt(mn)`.
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) = 4sqrt((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 sqrt((sin A)/(cos A))`
= 4 sin A. tan A ....(2)
Using equation (1) and equation (2) we get the required conditions.
Hence proved.
APPEARS IN
RELATED QUESTIONS
Prove the following trigonometric identities:
`(1 - cos^2 A) cosec^2 A = 1`
Prove the following trigonometric identities.
`1/(sec A - 1) + 1/(sec A + 1) = 2 cosec A cot A`
Prove the following identities:
cosec4 A (1 – cos4 A) – 2 cot2 A = 1
`(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))=1`
Write True' or False' and justify your answer the following :
The value of \[\sin \theta\] is \[x + \frac{1}{x}\] where 'x' is a positive real number .
Prove that:
`(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos^3 θ - sin^3 θ)/(cos θ - sin θ) = 2`
`5/(sin^2theta) - 5cot^2theta`, complete the activity given below.
Activity:
`5/(sin^2theta) - 5cot^2theta`
= `square (1/(sin^2theta) - cot^2theta)`
= `5(square - cot^2theta) ......[1/(sin^2theta) = square]`
= 5(1)
= `square`
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S = `square`
= `square/sintheta + sintheta/costheta`
= `(cos^2theta + sin^2theta)/square`
= `1/(sintheta*costheta)` ......`[cos^2theta + sin^2theta = square]`
= `1/sintheta xx 1/square`
= `square`
= R.H.S
If 5 sec θ – 12 cosec θ = 0, then find values of sin θ, sec θ
If tan θ = 3, then `(4 sin theta - cos theta)/(4 sin theta + cos theta)` is equal to ______.