Advertisements
Advertisements
Question
If 3 sin θ = 4 cos θ, then sec θ = ?
Advertisements
Solution
3 sin θ = 4 cos θ ...[Given]
∴ `(sin θ)/(cos θ) = 4/3`
∴ `tan θ = 4/3`
We know that,
1 + tan2θ = sec2θ
∴ `1 + (4/3)^2 = sec^2θ`
∴ `1 + 16/9 = sec^2θ`
∴ `sec^2θ = (9 + 16)/9`
∴ `sec^2θ = 25/9`
∴ `sec θ = 5/3` ...[Taking square root of both sides]
APPEARS IN
RELATED QUESTIONS
Prove the following trigonometric identities.
`(1 + cos A)/sin A = sin A/(1 - cos A)`
Prove the following trigonometric identities.
`(1 + cot A + tan A)(sin A - cos A) = sec A/(cosec^2 A) - (cosec A)/sec^2 A = sin A tan A - cos A cot A`
If 3 sin θ + 5 cos θ = 5, prove that 5 sin θ – 3 cos θ = ± 3.
`sin^2 theta + 1/((1+tan^2 theta))=1`
`1/((1+ sin θ)) + 1/((1 - sin θ)) = 2 sec^2 θ`
`sqrt((1-cos theta)/(1+cos theta)) = (cosec theta - cot theta)`
If \[sec\theta + tan\theta = x\] then \[tan\theta =\]
(cosec θ − sin θ) (sec θ − cos θ) (tan θ + cot θ) is equal to
Simplify
sin A `[[sinA -cosA],["cos A" " sinA"]] + cos A[[ cos A" sin A " ],[-sin A" cos A"]]`
Prove the following identity :
`cos^4A - sin^4A = 2cos^2A - 1`
Prove the following identity :
`cosA/(1 - tanA) + sinA/(1 - cotA) = sinA + cosA`
Prove the following identity :
`(tanθ + 1/cosθ)^2 + (tanθ - 1/cosθ)^2 = 2((1 + sin^2θ)/(1 - sin^2θ))`
Prove the following identity :
`(cosecθ)/(tanθ + cotθ) = cosθ`
Prove the following identity :
`[1/((sec^2θ - cos^2θ)) + 1/((cosec^2θ - sin^2θ))](sin^2θcos^2θ) = (1 - sin^2θcos^2θ)/(2 + sin^2θcos^2θ)`
Prove that `sin(90^circ - A).cos(90^circ - A) = tanA/(1 + tan^2A)`
Prove that : `(sin(90° - θ) tan(90° - θ) sec (90° - θ))/(cosec θ. cos θ. cot θ) = 1`
Without using trigonometric table, prove that
`cos^2 26° + cos 64° sin 26° + (tan 36°)/(cot 54°) = 2`
Prove the following identities.
`costheta/(1 + sintheta)` = sec θ – tan θ
If sin θ + cos θ = a and sec θ + cosec θ = b , then the value of b(a2 – 1) is equal to
Prove that `(cos^2θ)/(sinθ) + sin θ = "cosec" θ`.
