Advertisements
Advertisements
Question
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
Advertisements
Solution
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^2`
∴ tan2 θ = `625/49 - 1`
= `(625 - 49)/49`
= `576/49`
∴ tan θ = `24/7` ........(by taking square roots)
APPEARS IN
RELATED QUESTIONS
Prove that ` \frac{\sin \theta -\cos \theta +1}{\sin\theta +\cos \theta -1}=\frac{1}{\sec \theta -\tan \theta }` using the identity sec2 θ = 1 + tan2 θ.
Prove that (cosec A – sin A)(sec A – cos A) sec2 A = tan A.
Prove the following trigonometric identities.
`(1 + cos θ + sin θ)/(1 + cos θ - sin θ) = (1 + sin θ)/cos θ`
Prove the following identities:
`1/(cosA + sinA) + 1/(cosA - sinA) = (2cosA)/(2cos^2A - 1)`
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
`cos^2 theta /((1 tan theta))+ sin ^3 theta/((sin theta - cos theta))=(1+sin theta cos theta)`
Show that none of the following is an identity:
(i) `cos^2theta + cos theta =1`
Write the value of ` cosec^2 (90°- theta ) - tan^2 theta`
If \[\sin \theta = \frac{1}{3}\] then find the value of 9tan2 θ + 9.
If \[sec\theta + tan\theta = x\] then \[tan\theta =\]
Prove the following identity :
`sec^2A + cosec^2A = sec^2Acosec^2A`
Prove the following identity :
`cos^4A - sin^4A = 2cos^2A - 1`
Prove the following identity :
`(1 + tan^2A) + (1 + 1/tan^2A) = 1/(sin^2A - sin^4A)`
Prove the following identity :
`(1 + sinθ)/(cosecθ - cotθ) - (1 - sinθ)/(cosecθ + cotθ) = 2(1 + cotθ)`
Prove the following identity :
`(sec^2θ - sin^2θ)/tan^2θ = cosec^2θ - cos^2θ`
Without using trigonometric table , evaluate :
`(sin47^circ/cos43^circ)^2 - 4cos^2 45^circ + (cos43^circ/sin47^circ)^2`
If A = 60°, B = 30° verify that tan( A - B) = `(tan A - tan B)/(1 + tan A. tan B)`.
Prove that `cos θ/sin(90° - θ) + sin θ/cos (90° - θ) = 2`.
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
Proved that `(1 + secA)/secA = (sin^2A)/(1 - cos A)`.
