Advertisements
Advertisements
Question
Prove the following identity :
`2(sin^6θ + cos^6θ) - 3(sin^4θ + cos^4θ) + 1 = 0`
Advertisements
Solution
LHS = `2(sin^6θ + cos^6θ) - 3(sin^4θ + cos^4θ) + 1`
= `2(sin^6θ + cos^6θ) - 3(sin^4θ + cos^4θ) + 1`
= `2[(sin^2θ)^3 + (cos^2θ)^3] - 3(sin^4θ + cos^4θ) + 1`
= `2[(sin^2θ + cos^2θ){(sin^2θ)^2 + (cos^2θ)^2 - sin^2θcos^2θ}] - 3(sin^4θ + cos^4θ) + 1`
= `2{(sin^2θ)^2 + (cos^2θ)^2 - sin^2θcos^2θ} - 3(sin^4θ + cos^4θ) + 1`
= `2sin^4θ + 2cos^4θ - 2sin^2θcos^2θ - 3sin^4θ - 3cos^4θ + 1`
= `-sin^4θ - cos^4θ - 2sin^2θcos^2θ + 1`
= `-(sin^4θ + cos^4θ + 2sin^2θcos^2θ) + 1`
= `-(sin^2θ + cos^2θ)^2 + 1 = -1 + 1 = 0`
APPEARS IN
RELATED QUESTIONS
Show that `sqrt((1+cosA)/(1-cosA)) = cosec A + cot A`
Prove the following trigonometric identities.
`(cos theta - sin theta + 1)/(cos theta + sin theta - 1) = cosec theta + cot theta`
if `cosec theta - sin theta = a^3`, `sec theta - cos theta = b^3` prove that `a^2 b^2 (a^2 + b^2) = 1`
Prove that:
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
Prove the following identities:
sec4 A (1 – sin4 A) – 2 tan2 A = 1
Prove the following identity :
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
Prove the following identity :
`sqrt((secq - 1)/(secq + 1)) + sqrt((secq + 1)/(secq - 1))` = 2 cosesq
Without using trigonometric table , evaluate :
`(sin49^circ/sin41^circ)^2 + (cos41^circ/sin49^circ)^2`
Prove that `(cot "A" + "cosec A" - 1)/(cot "A" - "cosec A" + 1) = (1 + cos "A")/sin "A"`
Prove that cot2θ – tan2θ = cosec2θ – sec2θ.
