Advertisements
Advertisements
Question
Prove the following identity :
`(cosA + sinA)^2 + (cosA - sinA)^2 = 2`
Advertisements
Solution
LHS = `(cosA + sinA)^2 + (cosA - sinA)^2`
= `cos^2A + sin^2A + 2cosA.sinA + cos^2A + sin^2A - 2cosA.sinA`
= `2(cos^2A + sin^2A) = 2` = RHS
APPEARS IN
RELATED QUESTIONS
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(tan theta)/(1-cot theta) + (cot theta)/(1-tan theta) = 1+secthetacosectheta`
[Hint: Write the expression in terms of sinθ and cosθ]
If cos θ + cot θ = m and cosec θ – cot θ = n, prove that mn = 1
`(1+ tan theta + cot theta )(sintheta - cos theta) = ((sec theta)/ (cosec^2 theta)-( cosec theta)/(sec^2 theta))`
Prove the following identity :
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
Evaluate:
`(tan 65^circ)/(cot 25^circ)`
Without using the trigonometric table, prove that
tan 10° tan 15° tan 75° tan 80° = 1
Prove that cot2θ × sec2θ = cot2θ + 1.
If 1 + sin2θ = 3 sin θ cos θ, then prove that tan θ = 1 or `1/2`.
Prove the following:
(sin α + cos α)(tan α + cot α) = sec α + cosec α
Prove that `(1 + tan^2 A)/(1 + cot^2 A)` = sec2 A – 1
