Advertisements
Advertisements
Question
Prove the following trigonometric identities.
sec A (1 − sin A) (sec A + tan A) = 1
Advertisements
Solution
We have to prove sec A(1 − sin A)(sec A + tan A) = 1
We know that sec2 A − tan2 A − 1
So,
sec A(1 − sin A)(sec A + tan A) = {sec A(1 − sin A)}(sec A + tan A)
= (sec A − sec A sin A)(sec A + tan A)
= `(sec A - 1/cos A sin A) (sec A + tan A)` ...`(∵ sec theta = 1/costheta)`
= `(sec A - sin A/cos A) (sec A + tan A)` ...`(∵ tan theta = sin theta/costheta)`
= (sec A − tan A)(sec A + tan A)
= sec2 A − tan2 A
= 1 = R.H.S. ... (∵ sec2 θ = 1 tan2 θ)
APPEARS IN
RELATED QUESTIONS
Prove the following trigonometric identities.
(cosec θ − sec θ) (cot θ − tan θ) = (cosec θ + sec θ) ( sec θ cosec θ − 2)
If tan A = n tan B and sin A = m sin B , prove that `cos^2 A = ((m^2-1))/((n^2 - 1))`
Write the value of tan1° tan 2° ........ tan 89° .
\[\frac{1 + \tan^2 A}{1 + \cot^2 A}\]is equal to
Prove that:
`(cot A - 1)/(2 - sec^2 A) = cot A/(1 + tan A)`
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
Prove that `((1 - cos^2 θ)/cos θ)((1 - sin^2θ)/(sin θ)) = 1/(tan θ + cot θ)`
Prove that cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
If 1 + sin2θ = 3 sin θ cos θ, then prove that tan θ = 1 or `1/2`.
