Advertisements
Advertisements
प्रश्न
Prove that `(cot "A" + "cosec A" - 1)/(cot "A" - "cosec A" + 1) = (1 + cos "A")/sin "A"`
Advertisements
उत्तर
LHS = `(cot "A" + "cosec A" - 1)/(cot "A" - "cosec A" + 1)`
= `((cot "A" + "cosec A") - ("cosec"^2 "A" - cot^2 "A"))/(cot "A" - "cosec A" + 1)`
= `((cot "A" + "cosec A")("cosec A" + cot "A")("cosec A" - cot "A"))/(cot "A" - "cosec A" + 1)`
= `((cot "A" + "cosec A") [1 - "cosec A" - cot "A"])/(cot"A"-"cosec A"+1)`
= `((cot "A" + "cosec A") (1-"cosec A"+cot"A"))/(1-"cosec A"+cot"A")`
= cot A + cosec A
= `cos"A"/sin"A"+1/sin"A"=(cos"A"+1)/sin"A"`
= `(1+cos"A")/sin"A"`
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identity.
`cos^2 A + 1/(1 + cot^2 A) = 1`
Prove the following trigonometric identities.
`(cot^2 A(sec A - 1))/(1 + sin A) = sec^2 A ((1 - sin A)/(1 + sec A))`
If cos θ + cos2 θ = 1, prove that sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2 = 1
`1 + (tan^2 θ)/((1 + sec θ)) = sec θ`
If `cos theta = 2/3 , "write the value of" ((sec theta -1))/((sec theta +1))`
Write the value of \[\cot^2 \theta - \frac{1}{\sin^2 \theta}\]
Prove the following identity :
`(cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)`
Prove the following identities.
`(cot theta - cos theta)/(cot theta + cos theta) = ("cosec" theta - 1)/("cosec" theta + 1)`
sec 60° = ?
Show that tan 7° × tan 23° × tan 60° × tan 67° × tan 83° = `sqrt(3)`.
