Advertisements
Advertisements
प्रश्न
(sec θ + tan θ) . (sec θ – tan θ) = ?
उत्तर
(sec θ + tan θ)(sec θ – tan θ)
= sec2θ – tan2θ ......[∵ (a + b)(a – b) = a2 – b2]
= 1 ......`[(because 1 + tan^2theta = sec^2theta),(therefore sec^2theta - tan^2theta = 1)]`
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
if cos A + cos2 A = 1, prove that sin2 A + sin4 A = 1
Prove that `(sec theta - 1)/(sec theta + 1) = ((sin theta)/(1 + cos theta))^2`
Prove the following identities:
(1 + cot A – cosec A)(1 + tan A + sec A) = 2
Prove that:
2 sin2 A + cos4 A = 1 + sin4 A
Prove the following identities:
`(sinA - cosA + 1)/(sinA + cosA - 1) = cosA/(1 - sinA)`
`(cos ec^theta + cot theta )/( cos ec theta - cot theta ) = (cosec theta + cot theta )^2 = 1+2 cot^2 theta + 2cosec theta cot theta`
`(sin theta)/((sec theta + tan theta -1)) + cos theta/((cosec theta + cot theta -1))=1`
`(cot^2 theta ( sec theta - 1))/((1+ sin theta))+ (sec^2 theta(sin theta-1))/((1+ sec theta))=0`
Write True' or False' and justify your answer the following :
The value of the expression \[\sin {80}^° - \cos {80}^°\]
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
If a cos θ + b sin θ = 4 and a sin θ − b sin θ = 3, then a2 + b2 =
Prove the following identity :
secA(1 + sinA)(secA - tanA) = 1
Prove the following identity :
`(cosecA - sinA)(secA - cosA)(tanA + cotA) = 1`
Prove the following identity :
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
If tanA + sinA = m and tanA - sinA = n , prove that (`m^2 - n^2)^2` = 16mn
Prove that sin4A – cos4A = 1 – 2cos2A
Prove that sec2θ – cos2θ = tan2θ + sin2θ
`sqrt((1 - cos^2theta) sec^2 theta) = tan theta`
If sin A = `1/2`, then the value of sec A is ______.
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
