Advertisements
Advertisements
Question
Prove the following identities:
sec4 A (1 – sin4 A) – 2 tan2 A = 1
Advertisements
Solution
sec4 A (1 – sin4 A) – 2 tan2 A
= sec4 A – sec4 A sin4 A – 2 tan2 A
= `(sec^2A)^2 - 1/(cos^4A)sin^4A - 2tan^2A`
= (1 + tan2 A)2 – tan4 A – 2 tan2 A ...`[(sec^2A - tan^2A = 1), (sec^2A = 1 + tan^2A)]`
= (1)2 + (tan2 A)2 – 2 × 1 × tan2 A – tan4 A – 2 tan2 A
= 1 + tan4 A + 2 tan2 A – tan4 A – 2 tan2 A
= 1
APPEARS IN
RELATED QUESTIONS
if `cos theta = 5/13` where `theta` is an acute angle. Find the value of `sin theta`
Prove the following trigonometric identities.
tan2 A sec2 B − sec2 A tan2 B = tan2 A − tan2 B
Prove the following identities:
`(1 - cosA)/sinA + sinA/(1 - cosA)= 2cosecA`
Prove the following identities:
`cotA/(1 - tanA) + tanA/(1 - cotA) = 1 + tanA + cotA`
Prove that:
`sqrt(sec^2A + cosec^2A) = tanA + cotA`
If `secθ = 25/7 ` then find tanθ.
What is the value of (1 + cot2 θ) sin2 θ?
If cosθ = `5/13`, then find sinθ.
If x = h + a cos θ, y = k + b sin θ.
Prove that `((x - h)/a)^2 + ((y - k)/b)^2 = 1`.
Find the value of sin2θ + cos2θ
Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
