Advertisements
Advertisements
प्रश्न
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
उत्तर
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.
APPEARS IN
संबंधित प्रश्न
Prove that: `sqrt((sec theta - 1)/(sec theta + 1)) + sqrt((sec theta + 1)/(sec theta - 1)) = 2 cosec theta`
Prove that
`sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ)) = 2 sec θ`
If tanθ `= 3/4` then find the value of secθ.
(cosec θ − sin θ) (sec θ − cos θ) (tan θ + cot θ) is equal to
Prove the following identities:
`(sec"A"-1)/(sec"A"+1)=(sin"A"/(1+cos"A"))^2`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`cos 63^circ sec(90^circ - θ) = 1`
Prove that cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.
Prove that `cos θ/sin(90° - θ) + sin θ/cos (90° - θ) = 2`.
Prove that sec2θ + cosec2θ = sec2θ × cosec2θ.
(sec2 θ – 1) (cosec2 θ – 1) is equal to ______.
