Advertisements
Advertisements
Questions
Prove the following identities.
cot θ + tan θ = sec θ cosec θ
Prove that: tan θ + cot θ = sec θ cosec θ
Advertisements
Solution
L.H.S. = cot θ + tan θ
L.H.S. = `costheta/sintheta + sintheta/costheta`
L.H.S. = `(cos^2theta + sin^2theta)/(sintheta costheta)`
[cos2 θ + sin2 θ = 1]
L.H.S. = `1/(sintheta costheta)`
Use Reciprocal Identities:
The expression can be split into `(1/sin θ) xx (1/cos θ)`.
`1/sin θ` = cosec θ
`1/cos θ` = sec θ
L.H.S. = cosec θ.sec θ
L.H.S. = sec θ.cosec θ
∴ L.H.S. = R.H.S.
RELATED QUESTIONS
If `sec alpha=2/sqrt3` , then find the value of `(1-cosecalpha)/(1+cosecalpha)` where α is in IV quadrant.
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.
(1 + tan θ + sec θ) (1 + cot θ − cosec θ) = ______.
Prove the following trigonometric identities.
`(cos^2 theta)/sin theta - cosec theta + sin theta = 0`
Prove the following identities:
`(1+ sin A)/(cosec A - cot A) - (1 - sin A)/(cosec A + cot A) = 2(1 + cot A)`
If sec A + tan A = p, show that:
`sin A = (p^2 - 1)/(p^2 + 1)`
Write the value of ` sin^2 theta cos^2 theta (1+ tan^2 theta ) (1+ cot^2 theta).`
The value of sin2 29° + sin2 61° is
Prove the following identity :
`(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA) = 2`
prove that `1/(1 + cos(90^circ - A)) + 1/(1 - cos(90^circ - A)) = 2cosec^2(90^circ - A)`
Evaluate:
`(tan 65°)/(cot 25°)`
Prove the following identities.
`sqrt((1 + sin theta)/(1 - sin theta)` = sec θ + tan θ
If cot θ + tan θ = x and sec θ – cos θ = y, then prove that `(x^2y)^(2/3) – (xy^2)^(2/3)` = 1
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)
Prove that `1/("cosec" θ - cot θ) = "cosec" θ + cot θ`.
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 4 tanβ = 3, then `(4sinbeta-3cosbeta)/(4sinbeta+3cosbeta)=` ______.
Eliminate θ if x = r cosθ and y = r sinθ.
`(cos^2 θ)/(sin^2 θ) - 1/(sin^2 θ)`, in simplified form, is ______.
(1 – cos2 A) is equal to ______.
