Advertisements
Advertisements
प्रश्न
Prove the following identity:
(sin2θ – 1)(tan2θ + 1) + 1 = 0
उत्तर
L.H.S. = (sin2θ – 1)(tan2θ + 1) + 1
= (– cos2θ) sec2θ + 1
= `- cos^2θ xx 1/(cos^2θ) + 1`
= – 1 + 1
= 0
= R.H.S.
Hence Proved.
APPEARS IN
संबंधित प्रश्न
Prove the following identities:
`(i) cos4^4 A – cos^2 A = sin^4 A – sin^2 A`
`(ii) cot^4 A – 1 = cosec^4 A – 2cosec^2 A`
`(iii) sin^6 A + cos^6 A = 1 – 3sin^2 A cos^2 A.`
Evaluate without using trigonometric tables:
`cos^2 26^@ + cos 64^@ sin 26^@ + (tan 36^@)/(cot 54^@)`
Prove the following trigonometric identities
cosec6θ = cot6θ + 3 cot2θ cosec2θ + 1
` tan^2 theta - 1/( cos^2 theta )=-1`
`sqrt((1-cos theta)/(1+cos theta)) = (cosec theta - cot theta)`
Simplify : 2 sin30 + 3 tan45.
Prove the following identity :
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
Choose the correct alternative:
1 + cot2θ = ?
If 4 tanβ = 3, then `(4sinbeta-3cosbeta)/(4sinbeta+3cosbeta)=` ______.
`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.
