Advertisements
Advertisements
प्रश्न
If cos θ + cot θ = m and cosec θ – cot θ = n, prove that mn = 1
उत्तर
LHS = mn
`= (cosec theta + cot theta) (cosec theta - cot theta)`
`= cosece^2 theta - cot^2 theta`
= 1 [∵ `(1 + b)(a - b) = a^2 - b^2 cosec^2 theta - cot^2 theta = 1`]
=RHS
APPEARS IN
संबंधित प्रश्न
If sinθ + cosθ = p and secθ + cosecθ = q, show that q(p2 – 1) = 2p
Evaluate
`(sin ^2 63^@ + sin^2 27^@)/(cos^2 17^@+cos^2 73^@)`
The angles of depression of two ships A and B as observed from the top of a light house 60 m high are 60° and 45° respectively. If the two ships are on the opposite sides of the light house, find the distance between the two ships. Give your answer correct to the nearest whole number.
Prove that (cosec A – sin A)(sec A – cos A) sec2 A = tan A.
Prove the following trigonometric identities.
`sin^2 A + 1/(1 + tan^2 A) = 1`
Prove the following trigonometric identities.
`(1 - cos theta)/sin theta = sin theta/(1 + cos theta)`
Prove the following trigonometric identities.
`(tan^2 A)/(1 + tan^2 A) + (cot^2 A)/(1 + cot^2 A) = 1`
Prove the following trigonometric identities.
`[tan θ + 1/cos θ]^2 + [tan θ - 1/cos θ]^2 = 2((1 + sin^2 θ)/(1 - sin^2 θ))`
`(sec theta + tan theta )/( sec theta - tan theta ) = ( sec theta + tan theta )^2 = 1+2 tan^2 theta + 25 sec theta tan theta `
Write the value of `( 1- sin ^2 theta ) sec^2 theta.`
If `sec theta + tan theta = x," find the value of " sec theta`
Prove the following identity :
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
Prove the following identity :
`sqrt((secq - 1)/(secq + 1)) + sqrt((secq + 1)/(secq - 1))` = 2 cosesq
Prove that `sqrt((1 + sin A)/(1 - sin A))` = sec A + tan A.
Prove that `tan A/(1 + tan^2 A)^2 + cot A/(1 + cot^2 A)^2 = sin A.cos A`
Prove that `1/("cosec" theta - cot theta)` = cosec θ + cot θ
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
Show that tan 7° × tan 23° × tan 60° × tan 67° × tan 83° = `sqrt(3)`
If 1 + sin2θ = 3sinθ cosθ, then prove that tanθ = 1 or `1/2`.
If cosA + cos2A = 1, then sin2A + sin4A = 1.
