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`(1+ Cos Theta)(1- Costheta )(1+Cos^2 Theta)=1` - Mathematics

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प्रश्न

`(1+ cos theta)(1- costheta )(1+cos^2 theta)=1`

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उत्तर

LHS = `(1+costheta )(1-cos theta)(1+ cot^2 theta)`

       =` (1-cos^2 theta) cosec^2 theta`

       =` sin^2 theta xx cosec^2 theta`

       =` sin^2 theta xx1/(sin^2 theta)`

      =1

     = RHS

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अध्याय 8: Trigonometric Identities - Exercises 1

APPEARS IN

आर.एस. अग्रवाल Mathematics [English] Class 10
अध्याय 8 Trigonometric Identities
Exercises 1 | Q 4.1

संबंधित प्रश्न

Prove the following identities, where the angles involved are acute angles for which the expressions are defined:

`(cos A-sinA+1)/(cosA+sinA-1)=cosecA+cotA ` using the identity cosec2 A = 1 cot2 A.


Prove the following trigonometric identities.

`cosec theta sqrt(1 - cos^2 theta) = 1`


Prove the following trigonometric identities.

`tan theta - cot theta = (2 sin^2 theta - 1)/(sin theta cos theta)`


Prove the following trigonometric identities.

`sqrt((1 - cos A)/(1 + cos A)) = cosec A - cot A`


if `x/a cos theta + y/b sin theta = 1` and `x/a sin theta - y/b cos theta = 1` prove that `x^2/a^2 + y^2/b^2  = 2`


If cos θ + cot θ = m and cosec θ – cot θ = n, prove that mn = 1


Prove that:

(sec A − tan A)2 (1 + sin A) = (1 − sin A)


`(sec^2 theta-1) cot ^2 theta=1`


`1+(tan^2 theta)/((1+ sec theta))= sec theta`


cosec4θ − cosec2θ = cot4θ + cot2θ


`(1-tan^2 theta)/(cot^2-1) = tan^2 theta`


Write the value of `(1+ tan^2 theta ) ( 1+ sin theta ) ( 1- sin theta)`


Find the value of ( sin2 33° + sin2 57°).


Prove that `(sin (90° - θ))/cos θ + (tan (90° - θ))/cot θ + (cosec (90° - θ))/sec θ = 3`.


Prove that:

`(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(2 sin^2 A - 1)`


The value of sin2θ + `1/(1 + tan^2 theta)` is equal to 


Prove that `(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ)2 


Prove that cosec θ – cot θ = `sin theta/(1 + cos theta)`


If cosec A – sin A = p and sec A – cos A = q, then prove that `("p"^2"q")^(2/3) + ("pq"^2)^(2/3)` = 1


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


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