Advertisements
Advertisements
Questions
Prove the following trigonometric identities.
sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
Prove the following:
sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
Advertisements
Solution
We need to prove `sec^6 theta = tan^6 theta + 3 tan^2 theta sec^2 theta + 1`
Solving the L.H.S, we get
`sec^6 theta = (sec^2 theta)^3`
`= (1 + tan^2 theta)^3`
Further using the identity `(a + b)^3 = a^3 + b^3 + 3a^2b + 3ab^2`, we get
`(1 + tan^2 theta)^3 = 1 + tan^6 theta + 3(1)^2 (tan^2 theta) + 3(1)(tan^2 theta)^2`
`= 1 + tan^6 theta + 3 tan^2 theta + 3 tan^4 theta`
`= 1 + tan^6 theta + 3 tan^2 theta + 3 tan^4 theta`
`= 1 + tan^6 theta + 3 tan^2 theta (1 + tan^2 theta)`
`= 1 + tan^6 theta + 3 tan^2 theta sec^2 theta` (using `1 + tan^2 theta = sec^2 theta`)
Hence proved.
RELATED QUESTIONS
Prove the following identities:
`(1 + sinA)/cosA + cosA/(1 + sinA) = 2secA`
Prove the following identities:
`sqrt((1 - sinA)/(1 + sinA)) = cosA/(1 + sinA)`
Find the value of ` ( sin 50°)/(cos 40°)+ (cosec 40°)/(sec 50°) - 4 cos 50° cosec 40 °`
Simplify : 2 sin30 + 3 tan45.
For ΔABC , prove that :
`sin((A + B)/2) = cos"C/2`
Prove that:
`sqrt(( secθ - 1)/(secθ + 1)) + sqrt((secθ + 1)/(secθ - 1)) = 2 "cosec"θ`
If 5x = sec θ and `5/x` = tan θ, then `x^2 - 1/x^2` is equal to
Prove that `(cos(90^circ - A))/(sin A) = (sin(90^circ - A))/(cos A)`.
Prove that `(1 + sin θ)/(1 - sin θ) = (sec θ + tan θ)^2`.
If cosec A – sin A = p and sec A – cos A = q, then prove that `(p^2q)^(2/3) + (pq^2)^(2/3) = 1`.
