Advertisements
Advertisements
Question
Prove the following identities.
sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
Advertisements
Solution
sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
L.H.S = sec6 θ
= (sec2 θ)3 = (1 + tan2 θ)3
= 1 + (tan2 θ)3 + 3 (1) (tan2 θ) (1 + tan2 θ) ......[(a + b)3 = a3 + b3 + 3 ab (a + b)]
= 1 + tan6 θ + 3 tan2 θ (1 + tan2 θ)
= 1 + tan6 θ + 3 tan2 θ (sec2 θ)
= 1 + tan6 θ + 3 tan2 θ sec2 θ
= tan6 θ + 3 tan2 θ sec2 θ + 1
L.H.S = R.H.S
APPEARS IN
RELATED QUESTIONS
Evaluate without using trigonometric tables:
`cos^2 26^@ + cos 64^@ sin 26^@ + (tan 36^@)/(cot 54^@)`
Prove the following trigonometric identities.
`1/(sec A - 1) + 1/(sec A + 1) = 2 cosec A cot A`
Prove the following identities:
`tan A - cot A = (1 - 2cos^2A)/(sin A cos A)`
If \[sec\theta + tan\theta = x\] then \[tan\theta =\]
Prove the following identity :
`tan^2θ/(tan^2θ - 1) + (cosec^2θ)/(sec^2θ - cosec^2θ) = 1/(sin^2θ - cos^2θ)`
Verify that the points A(–2, 2), B(2, 2) and C(2, 7) are the vertices of a right-angled triangle.
Express (sin 67° + cos 75°) in terms of trigonometric ratios of the angle between 0° and 45°.
`(sin A)/(1 + cos A) + (1 + cos A)/(sin A)` = 2 cosec A
Prove that:
`(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos^3 θ - sin^3 θ)/(cos θ - sin θ) = 2`
If sec θ + tan θ = `sqrt(3)`, complete the activity to find the value of sec θ – tan θ
Activity:
`square` = 1 + tan2θ ......[Fundamental trigonometric identity]
`square` – tan2θ = 1
(sec θ + tan θ) . (sec θ – tan θ) = `square`
`sqrt(3)*(sectheta - tan theta)` = 1
(sec θ – tan θ) = `square`
