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Question
If A + B = 90°, show that sec2 A + sec2 B = sec2 A. sec2 B.
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Solution
LHS = sec2A + sec2B
= `1/(cos^2A) + 1/(cos^2 B)`
= `1/(cos^2A) + 1/(cos^2(90° - A))`
= `1/(cos^2 A) + 1/(sin^2 A)`
= `1/( sin^ A. cos^2 A)`
= sec2 A cosec2 A
= sec2 A cosec2 (90° - B)
= sec2A. sec2 B = RHS
Hence proved.
RELATED QUESTIONS
If a cos θ + b sin θ = m and a sin θ – b cos θ = n, prove that a2 + b2 = m2 + n2
Prove the following identities:
(cosec A + sin A) (cosec A – sin A) = cot2 A + cos2 A
(i)` (1-cos^2 theta )cosec^2theta = 1`
` tan^2 theta - 1/( cos^2 theta )=-1`
`tan theta/(1+ tan^2 theta)^2 + cottheta/(1+ cot^2 theta)^2 = sin theta cos theta`
`(sin theta +cos theta )/(sin theta - cos theta)+(sin theta- cos theta)/(sin theta + cos theta) = 2/((sin^2 theta - cos ^2 theta)) = 2/((2 sin^2 theta -1))`
Write the value of `(1 + cot^2 theta ) sin^2 theta`.
Prove that sin θ sin( 90° - θ) - cos θ cos( 90° - θ) = 0
a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p2 – q2 is equal to
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.
