Advertisements
Advertisements
Question
Prove the following identities:
`(1+ sin A)/(cosec A - cot A) - (1 - sin A)/(cosec A + cot A) = 2(1 + cot A)`
Advertisements
Solution
L.H.S. = `(1 + sin A)/(cosec A - cot A) - (1 - sin A)/(cosec A + cot A)`
= `((1 + sin A)(cosec A + cot A) - (1 - sin A)(cosec A - cot A))/((cosec A - cot A)(cosec A + cot A))`
= `(cosec A + cot A + sin A cosec A + sin A cot A - cosec A + cot A + sin A cosec A - sin A cos A)/(cosec^2A - cot^2A)`
= 2 cot A + 2 sin A cosec A
= 2 cot A + 2 `1/(cosec A) xx cosec A`
= 2 (cot A + 1)
Hence proved.
RELATED QUESTIONS
`(sec^2 theta-1) cot ^2 theta=1`
Write the value of ` cosec^2 (90°- theta ) - tan^2 theta`
What is the value of (1 − cos2 θ) cosec2 θ?
If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, then\[\frac{x^2}{a^2} + \frac{y^2}{b^2}\]
Without using trigonometric identity , show that :
`sin42^circ sec48^circ + cos42^circ cosec48^circ = 2`
Prove that:
`sqrt((sectheta - 1)/(sec theta + 1)) + sqrt((sectheta + 1)/(sectheta - 1)) = 2cosectheta`
Prove that `( 1 + sin θ)/(1 - sin θ) = 1 + 2 tan θ/cos θ + 2 tan^2 θ` .
`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`
If tan θ + cot θ = 2, then tan2θ + cot2θ = ?
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S = `square`
= `square/sintheta + sintheta/costheta`
= `(cos^2theta + sin^2theta)/square`
= `1/(sintheta*costheta)` ......`[cos^2theta + sin^2theta = square]`
= `1/sintheta xx 1/square`
= `square`
= R.H.S
