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प्रश्न
What is the value of (1 + cot2 θ) sin2 θ?
उत्तर
We have,
`(1+cot^2 θ)sin^2θ= cosec^2θxxsin^2θ`
`= (1/sinθ)^2 xx sin^2θ`
= `1/sin^2θxxsin^2θ`
`=1`
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संबंधित प्रश्न
Prove the following trigonometric identities.
`((1 + tan^2 theta)cot theta)/(cosec^2 theta) = tan theta`
Prove the following trigonometric identities.
(sec A − cosec A) (1 + tan A + cot A) = tan A sec A − cot A cosec A
Prove the following identities:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
Prove the following identities:
(cosec A + sin A) (cosec A – sin A) = cot2 A + cos2 A
Prove that:
(sec A − tan A)2 (1 + sin A) = (1 − sin A)
Show that : `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec A cosec A`
If sin A + cos A = p and sec A + cosec A = q, then prove that : q(p2 – 1) = 2p.
Prove that:
(cosec A – sin A) (sec A – cos A) sec2 A = tan A
`(sec^2 theta-1) cot ^2 theta=1`
`(1+ cos theta - sin^2 theta )/(sin theta (1+ cos theta))= cot theta`
From the figure find the value of sinθ.
Write True' or False' and justify your answer the following:
\[ \cos \theta = \frac{a^2 + b^2}{2ab}\]where a and b are two distinct numbers such that ab > 0.
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
Prove the following identity :
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
Prove that `sqrt((1 + sin θ)/(1 - sin θ))` = sec θ + tan θ.
Prove that `sqrt((1 + cos A)/(1 - cos A)) = (tan A + sin A)/(tan A. sin A)`
sin4A – cos4A = 1 – 2cos2A. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` .....`[sin^2"A" + square = 1]`
= `square` – cos2A .....[sin2A = 1 – cos2A]
= `square`
= R.H.S
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
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
