Advertisements
Advertisements
Question
Prove the following identity :
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
Advertisements
Solution
LHS = `(sinA + sinB)/(cosA + cosB) + (cosA - cosB)/(sinA - sinB) `
= `((sinA + sinB)(sinA - sinB) + (cosA + cosB)(cosA - cosB))/((cosA + cosB)(sinA - sinB))`
= `(sin^2A - sin^2B + cos^2A - cos^2B)/((cosA + cosB)(sinA - sinB))`
= `((sin^2A + cos^2A) - (sin^2B + cos^2B))/((cosA + cosB)(sinA - sinB)`
= `(1-1)/((cosA + cosB)(sinA - sinB))`
= `0/((cosA + cosB)(sinA - sinB))`
= 0
`(sinA + sinB)/(cosA + cosB) + (cosA - cosB)/(sinA - sinB) = 0`
APPEARS IN
RELATED QUESTIONS
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(cosec θ – cot θ)^2 = (1-cos theta)/(1 + cos theta)`
Prove the following trigonometric identities.
sec A (1 − sin A) (sec A + tan A) = 1
Prove that:
(tan A + cot A) (cosec A – sin A) (sec A – cos A) = 1
`(1+ cos theta + sin theta)/( 1+ cos theta - sin theta )= (1+ sin theta )/(cos theta)`
If `( tan theta + sin theta ) = m and ( tan theta - sin theta ) = n " prove that "(m^2-n^2)^2 = 16 mn .`
prove that `1/(1 + cos(90^circ - A)) + 1/(1 - cos(90^circ - A)) = 2cosec^2(90^circ - A)`
Prove that identity:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
Choose the correct alternative:
cos θ. sec θ = ?
`(1 - tan^2 45^circ)/(1 + tan^2 45^circ)` = ?
Prove that `(tan(90 - theta) + cot(90 - theta))/("cosec" theta)` = sec θ
