Advertisements
Advertisements
प्रश्न
Prove the following identity :
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
उत्तर
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
संबंधित प्रश्न
Prove the following trigonometric identities.
(1 + tan2θ) (1 − sinθ) (1 + sinθ) = 1
Prove the following identities:
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
`(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))=1`
If tan A = n tan B and sin A = m sin B , prove that `cos^2 A = ((m^2-1))/((n^2 - 1))`
Prove the following identity :
`(secA - 1)/(secA + 1) = (1 - cosA)/(1 + cosA)`
Prove that: (1+cot A - cosecA)(1 + tan A+ secA) =2.
If a cos θ – b sin θ = c, then prove that (a sin θ + b cos θ) = `± sqrt("a"^2 + "b"^2 -"c"^2)`
If tan θ = `x/y`, then cos θ is equal to ______.
(1 – cos2 A) is equal to ______.
`1/sin^2θ - 1/cos^2θ - 1/tan^2θ - 1/cot^2θ - 1/sec^2θ - 1/("cosec"^2θ) = -3`, then find the value of θ.
