Advertisements
Advertisements
प्रश्न
`sqrt((1+sin theta)/(1-sin theta)) = (sec theta + tan theta)`
Advertisements
उत्तर
LHS = `sqrt((1+sin theta)/(1-sin theta))`
=`sqrt(((1+ sin theta))/(1- sin theta) xx ((1+sin theta))/(1+ sin theta))`
=` sqrt(((1+sin theta)^2)/(1-sin^2 theta))`
=`sqrt(((1+ sin theta)^2)/(cos^2 theta))`
=`(1+sin theta)/cos theta`
=`1/cos theta+ (sin theta)/(cos theta)`
= (sec 𝜃 + tan 𝜃)
= RHS
APPEARS IN
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(1+ secA)/sec A = (sin^2A)/(1-cosA)`
[Hint : Simplify LHS and RHS separately.]
Prove the following trigonometric identities.
`(1 + cos A)/sin^2 A = 1/(1 - cos A)`
Prove that `(sec theta - 1)/(sec theta + 1) = ((sin theta)/(1 + cos theta))^2`
Prove the following identities:
`((1 + tan^2A)cotA)/(cosec^2A) = tan A`
Prove the following identities:
(cosec A + sin A) (cosec A – sin A) = cot2 A + cos2 A
If sin A + cos A = m and sec A + cosec A = n, show that : n (m2 – 1) = 2 m
Prove that:
`(sinA - cosA)(1 + tanA + cotA) = secA/(cosec^2A) - (cosecA)/(sec^2A)`
`cot^2 theta - 1/(sin^2 theta ) = -1`a
Write the value of `(1+ tan^2 theta ) ( 1+ sin theta ) ( 1- sin theta)`
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
Prove the following identity:
`cosA/(1 + sinA) = secA - tanA`
Prove the following identity :
`(cosA + sinA)^2 + (cosA - sinA)^2 = 2`
Prove the following identity :
`(cotA - cosecA)^2 = (1 - cosA)/(1 + cosA)`
Prove the following identity :
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
Prove the following identity :
`[1/((sec^2θ - cos^2θ)) + 1/((cosec^2θ - sin^2θ))](sin^2θcos^2θ) = (1 - sin^2θcos^2θ)/(2 + sin^2θcos^2θ)`
Without using trigonometric table, prove that
`cos^2 26° + cos 64° sin 26° + (tan 36°)/(cot 54°) = 2`
Prove that sin4A – cos4A = 1 – 2cos2A
Prove that `sqrt(sec^2 theta + "cosec"^2 theta) = tan theta + cot theta`
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
If tan θ = `x/y`, then cos θ is equal to ______.
