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:
sec2θ + cosec2θ = sec2θ x cosec2θ
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(cos A-sinA+1)/(cosA+sinA-1)=cosecA+cotA ` using the identity cosec2 A = 1 cot2 A.
Prove the following trigonometric identities.
`((1 + tan^2 theta)cot theta)/(cosec^2 theta) = tan theta`
If x = r cos A cos B, y = r cos A sin B and z = r sin A, show that : x2 + y2 + z2 = r2
`cosec theta (1+costheta)(cosectheta - cot theta )=1`
`sin^2 theta + cos^4 theta = cos^2 theta + sin^4 theta`
`(1-tan^2 theta)/(cot^2-1) = tan^2 theta`
Show that none of the following is an identity:
(i) `cos^2theta + cos theta =1`
Prove that:
`(sin^2θ)/(cosθ) + cosθ = secθ`
Prove the following identity :
`(secθ - tanθ)^2 = (1 - sinθ)/(1 + sinθ)`
If x = acosθ , y = bcotθ , prove that `a^2/x^2 - b^2/y^2 = 1.`
If sec θ = `25/7`, then find the value of tan θ.
Find the value of ( sin2 33° + sin2 57°).
Prove that: `1/(sec θ - tan θ) = sec θ + tan θ`.
Prove the following identities.
`sqrt((1 + sin theta)/(1 - sin theta)) + sqrt((1 - sin theta)/(1 + sin theta))` = 2 sec θ
Which is not correct formula?
Prove that sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ.
If tan θ = `x/y`, then cos θ is equal to ______.
(1 – cos2 A) is equal to ______.
Prove that `(1 + tan^2 A)/(1 + cot^2 A)` = sec2 A – 1
