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Question
Prove that ( 1 + tan A)2 + (1 - tan A)2 = 2 sec2A
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Solution
LHS = ( 1 + tan A)2 + (1 - tan A)2
= 1 + 2 tan A + tan2A + 1 - 2 tan A + tan2A
= 2( 1 + tan2A)
= 2 sec2A
= RHS
Hence proved.
RELATED QUESTIONS
(secA + tanA) (1 − sinA) = ______.
Prove that `(tan^2 theta)/(sec theta - 1)^2 = (1 + cos theta)/(1 - cos theta)`
`(1-tan^2 theta)/(cot^2-1) = tan^2 theta`
Write the value of `(1 + tan^2 theta ) cos^2 theta`.
If `sec theta = x ,"write the value of tan" theta`.
What is the value of (1 + tan2 θ) (1 − sin θ) (1 + sin θ)?
Prove the following identity :
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
Prove that `sin^2 θ/ cos^2 θ + cos^2 θ/sin^2 θ = 1/(sin^2 θ. cos^2 θ) - 2`.
Prove that `costheta/(1 + sintheta) = (1 - sintheta)/(costheta)`
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
