Advertisements
Advertisements
Question
Show that tan4θ + tan2θ = sec4θ – sec2θ.
Advertisements
Solution
L.H.S = tan4θ + tan2θ
= tan2θ(tan2θ + 1)
= tan2θ.sec2θ ...[∵ sec2θ = tan2θ + 1]
= (sec2θ – 1).sec2θ ...[∵ tan2θ = sec2θ – 1]
= sec4θ – sec2θ
= R.H.S
APPEARS IN
RELATED QUESTIONS
Prove that `\frac{\sin \theta -\cos \theta }{\sin \theta +\cos \theta }+\frac{\sin\theta +\cos \theta }{\sin \theta -\cos \theta }=\frac{2}{2\sin^{2}\theta -1}`
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
`(sintheta - 2sin^3theta)/(2costheta - costheta) =tan theta`
Prove the following trigonometric identities.
`((1 + sin theta - cos theta)/(1 + sin theta + cos theta))^2 = (1 - cos theta)/(1 + cos theta)`
Prove the following trigonometric identities.
(1 + cot A − cosec A) (1 + tan A + sec A) = 2
Prove the following identities:
`(1 - sinA)/(1 + sinA) = (secA - tanA)^2`
`1/((1+ sin θ)) + 1/((1 - sin θ)) = 2 sec^2 θ`
`(sin theta)/((sec theta + tan theta -1)) + cos theta/((cosec theta + cot theta -1))=1`
What is the value of (1 − cos2 θ) cosec2 θ?
What is the value of \[6 \tan^2 \theta - \frac{6}{\cos^2 \theta}\]
What is the value of (1 + tan2 θ) (1 − sin θ) (1 + sin θ)?
If sec θ + tan θ = x, then sec θ =
\[\frac{\sin \theta}{1 + \cos \theta}\]is equal to
Prove the following identity :
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
Prove the following identity :
`(cotA - cosecA)^2 = (1 - cosA)/(1 + cosA)`
Evaluate:
sin2 34° + sin2 56° + 2 tan 18° tan 72° – cot2 30°
Prove that sin4θ - cos4θ = sin2θ - cos2θ
= 2sin2θ - 1
= 1 - 2 cos2θ
Prove that sec2 (90° - θ) + tan2 (90° - θ) = 1 + 2 cot2 θ.
Prove that:
`(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(2 sin^2 A - 1)`
Prove the following identities.
sec4 θ (1 – sin4 θ) – 2 tan2 θ = 1
Prove that sec2θ + cosec2θ = sec2θ × cosec2θ
