Advertisements
Advertisements
Question
Prove that ( 1 + tan A)2 + (1 - tan A)2 = 2 sec2A
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.
APPEARS IN
RELATED QUESTIONS
Prove the following trigonometric identities:
`(\text{i})\text{ }\frac{\sin \theta }{1-\cos \theta }=\text{cosec}\theta+\cot \theta `
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(1+ secA)/sec A = (sin^2A)/(1-cosA)`
[Hint : Simplify LHS and RHS separately.]
Prove the following identities:
`(costhetacottheta)/(1 + sintheta) = cosectheta - 1`
Prove the following identities:
cosec4 A (1 – cos4 A) – 2 cot2 A = 1
If 2 sin A – 1 = 0, show that: sin 3A = 3 sin A – 4 sin3 A
If tan A = n tan B and sin A = m sin B , prove that `cos^2 A = ((m^2-1))/((n^2 - 1))`
If a cot θ + b cosec θ = p and b cot θ − a cosec θ = q, then p2 − q2
Prove the following identity :
`tanA - cotA = (1 - 2cos^2A)/(sinAcosA)`
Prove the following identity :
`(1 + cotA + tanA)(sinA - cosA) = secA/(cosec^2A) - (cosecA)/sec^2A`
Without using the trigonometric table, prove that
cos 1°cos 2°cos 3° ....cos 180° = 0.