Advertisements
Advertisements
प्रश्न
Show that : tan 10° tan 15° tan 75° tan 80° = 1
Advertisements
उत्तर
L.H.S. = tan 10° tan 15° tan 75° tan 80°
= tan(90° – 80°) tan(90° – 75°) tan 75° tan 80°
= cot 80° cot 75° tan 75° tan 80° ...[∵ tan(90° – θ] = cot θ]]
= tan 80° cot 80° × tan 75° cot 75°
= 1 × 1
= 1 = R.H.S. ...(∵ tan A cot A = 1)
संबंधित प्रश्न
Prove the following identities:
`(i) (sinθ + cosecθ)^2 + (cosθ + secθ)^2 = 7 + tan^2 θ + cot^2 θ`
`(ii) (sinθ + secθ)^2 + (cosθ + cosecθ)^2 = (1 + secθ cosecθ)^2`
`(iii) sec^4 θ– sec^2 θ = tan^4 θ + tan^2 θ`
If sinθ + sin2 θ = 1, prove that cos2 θ + cos4 θ = 1
If `(x/a sin a - y/b cos theta) = 1 and (x/a cos theta + y/b sin theta ) =1, " prove that "(x^2/a^2 + y^2/b^2 ) =2`
Prove the following identity :
`cosecA + cotA = 1/(cosecA - cotA)`
Prove the following identity :
`(tanθ + 1/cosθ)^2 + (tanθ - 1/cosθ)^2 = 2((1 + sin^2θ)/(1 - sin^2θ))`
Prove that:
`sqrt((sectheta - 1)/(sec theta + 1)) + sqrt((sectheta + 1)/(sectheta - 1)) = 2cosectheta`
If A = 30°, verify that `sin 2A = (2 tan A)/(1 + tan^2 A)`.
Choose the correct alternative:
cos θ. sec θ = ?
Prove that cos2θ . (1 + tan2θ) = 1. Complete the activity given below.
Activity:
L.H.S = `square`
= `cos^2theta xx square .....[1 + tan^2theta = square]`
= `(cos theta xx square)^2`
= 12
= 1
= R.H.S
If 4 tanβ = 3, then `(4sinbeta-3cosbeta)/(4sinbeta+3cosbeta)=` ______.
