Advertisements
Advertisements
Question
Prove the following identity :
`(1 - tanA)^2 + (1 + tanA)^2 = 2sec^2A`
Advertisements
Solution
LHS = `(1 - tanA)^2 + (1 + tanA)^2`
= `1 + tan^2A - 2tanA + 1 + tan^2A + 2tanA`
= `2(1 + tan^2A) = 2sec^2A` = RHS
APPEARS IN
RELATED QUESTIONS
If m=(acosθ + bsinθ) and n=(asinθ – bcosθ) prove that m2+n2=a2+b2
`(1+ cos theta)(1- costheta )(1+cos^2 theta)=1`
Define an identity.
Write the value of \[\cot^2 \theta - \frac{1}{\sin^2 \theta}\]
The value of \[\sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}}\]
Simplify
sin A `[[sinA -cosA],["cos A" " sinA"]] + cos A[[ cos A" sin A " ],[-sin A" cos A"]]`
For ΔABC , prove that :
`tan ((B + C)/2) = cot "A/2`
If cosec θ + cot θ = p, then prove that cos θ = `(p^2 - 1)/(p^2 + 1)`
If 2 cos θ + sin θ = `1(θ ≠ π/2)`, then 7 cos θ + 6 sin θ is equal to ______.
Prove that `(1 + tan^2 A)/(1 + cot^2 A)` = sec2 A – 1
