Advertisements
Advertisements
Question
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
Options
2 tan θ
2 sec θ
2 cosec θ
2 tan θ sec θ
Advertisements
Solution
The given expression is `tan θ /(secθ-1)+tan θ/(sec θ+1)`
=` (tan θ (sec θ+1)+tan θ(secθ-1))/((secθ-1)(secθ+1))`
= `(tan θ sec θ+tanθ+tan θ secθ-tan θ)/(sec^2θ-1)`
=`( 2tanθ secθ)/tan^2θ`
=`(2secθ)/tan θ`
= `(2 1/cos θ)/(sinθ/cos θ)`
=`2 1/ sinθ`
= `2 cosec θ`
APPEARS IN
RELATED QUESTIONS
`"If "\frac{\cos \alpha }{\cos \beta }=m\text{ and }\frac{\cos \alpha }{\sin \beta }=n " show that " (m^2 + n^2 ) cos^2 β = n^2`
`(1+tan^2A)/(1+cot^2A)` = ______.
Prove the following trigonometric identities.
`"cosec" theta sqrt(1 - cos^2 theta) = 1`
Prove the following trigonometric identities.
`(cos^2 theta)/sin theta - cosec theta + sin theta = 0`
Prove the following trigonometric identities.
`(1 + sin θ)/cos θ+ cos θ/(1 + sin θ) = 2 sec θ`
Prove the following trigonometric identities.
`cos A/(1 - tan A) + sin A/(1 - cot A) = sin A + cos A`
Prove the following identities:
`(1 + sin A)/(1 - sin A) = (cosec A + 1)/(cosec A - 1)`
Prove the following identities:
(1 – tan A)2 + (1 + tan A)2 = 2 sec2A
Prove that:
(1 + tan A . tan B)2 + (tan A – tan B)2 = sec2 A sec2 B
Prove the following identities:
`((cosecA - cotA)^2 + 1)/(secA(cosecA - cotA)) = 2cotA`
If` (sec theta + tan theta)= m and ( sec theta - tan theta ) = n ,` show that mn =1
Write the value of `(1 + cot^2 theta ) sin^2 theta`.
If \[\sin \theta = \frac{1}{3}\] then find the value of 9tan2 θ + 9.
If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, then\[\frac{x^2}{a^2} + \frac{y^2}{b^2}\]
Prove the following identity :
tanA+cotA=secAcosecA
Prove the following identity :
`(tanθ + 1/cosθ)^2 + (tanθ - 1/cosθ)^2 = 2((1 + sin^2θ)/(1 - sin^2θ))`
prove that `1/(1 + cos(90^circ - A)) + 1/(1 - cos(90^circ - A)) = 2cosec^2(90^circ - A)`
Choose the correct alternative:
cos θ. sec θ = ?
Choose the correct alternative:
1 + cot2θ = ?
`5/(sin^2theta) - 5cot^2theta`, complete the activity given below.
Activity:
`5/(sin^2theta) - 5cot^2theta`
= `square (1/(sin^2theta) - cot^2theta)`
= `5(square - cot^2theta) ......[1/(sin^2theta) = square]`
= 5(1)
= `square`
