Advertisements
Advertisements
Question
Prove the following identities:
`(1 + sinA)/cosA + cosA/(1 + sinA) = 2secA`
Advertisements
Solution
L.H.S. = `(1 + sinA)/cosA + cosA/(1 + sinA)`
= `((1 + sinA)^2 + cos^2A)/(cosA(1 + sinA))`
= `(1 + sin^2A + 2sinA + cos^2A)/(cosA(1 + sinA))`
= `(1 + 2sinA + 1)/(cosA(1 + sinA))`
= `(2(1 + sinA))/(cosA(1 + sinA))`
= 2 sec A = R.H.S.
APPEARS IN
RELATED QUESTIONS
Without using trigonometric tables evaluate
`(sin 35^@ cos 55^@ + cos 35^@ sin 55^@)/(cosec^2 10^@ - tan^2 80^@)`
Prove the following identities:
`(costhetacottheta)/(1 + sintheta) = cosectheta - 1`
Prove that:
`(tanA + 1/cosA)^2 + (tanA - 1/cosA)^2 = 2((1 + sin^2A)/(1 - sin^2A))`
Write the value of `cosec^2 theta (1+ cos theta ) (1- cos theta).`
If sin2 θ cos2 θ (1 + tan2 θ) (1 + cot2 θ) = λ, then find the value of λ.
Prove the following identity :
secA(1 + sinA)(secA - tanA) = 1
Prove the following identities:
`(sec"A"-1)/(sec"A"+1)=(sin"A"/(1+cos"A"))^2`
Prove that: 2(sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = 0.
Prove that the following identities:
Sec A( 1 + sin A)( sec A - tan A) = 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.
