हिंदी

Prove that 1+secθ-tanθ1+secθ+tanθ=1-sinθcosθ

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प्रश्न

Prove that `(1 + sec theta - tan theta)/(1 + sec theta + tan theta) = (1 - sin theta)/cos theta`

योग
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उत्तर

L.H.S = `(1 + sec theta - tan theta)/(1 + sec theta + tan theta)`

= `(1 + 1//cos theta - sin theta//cos theta)/(1 + 1//cos theta + sin theta//cos theta)`  ...`[∵ sec theta = 1/cos theta and tan theta = sin theta/cos theta]`

= `(cos theta + 1 - sin theta)/(cos theta + 1 + sin theta)`

= `((cos theta + 1) - sin theta)/((cos theta + 1) + sin theta)`

= `(2 cos^2  theta/2 - 2 sin  theta/2 * cos  theta/2)/(2 cos^2  theta/2 + 2 sin  theta/2 * cos  theta/2)`  ...`[∵ 1 + cos theta = 2 cos^2  theta/2 and sin theta = 2sin  theta/2 cos  theta/2]`

= `(2cos^2  theta/2 - 2 sin  theta/2 * cos  theta/2)/(2cos^2  theta/2 + 2sin  theta/2 * cos  theta/2)`

= `(2cos  theta/2 (cos  theta/2 - sin  theta/2))/(2cos  theta/2(cos  theta/2 + sin  theta/2))`

= `(cos  theta/2 - sin  theta/2)/(cos  theta/2 + sin  theta/2) xx ((cos  theta/2 - sin  theta/2))/((cos  theta/2 - sin  theta/2))`  ...[By rationalisation]

= `(cos  theta/2 - sin  theta/2)^2/((cos^2  theta/2 - sin^2  theta/2))`  ...[∵ (a – b)2 = a2 + b2 – 2ab and (a – b)(a + b) = (a2 – b2)]

= `((cos^2  theta/2 + sin^2  theta/2) - (2 sin  theta/2 * cos  theta/2))/cos theta`   ...`[∵ cos^2  theta/2 - sin^2  theta/2 = cos theta]`

= `(1 - sin theta)/cos theta`   ...`[∵ sin^2  theta/2 + cos^2  theta/2 = 1]`

= R.H.S

Hence proved.

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अध्याय 8: Introduction To Trigonometry and Its Applications - Exercise 8.4 [पृष्ठ ९९]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 10
अध्याय 8 Introduction To Trigonometry and Its Applications
Exercise 8.4 | Q 12 | पृष्ठ ९९

संबंधित प्रश्न

If cosθ + sinθ = √2 cosθ, show that cosθ – sinθ = √2 sinθ.


Prove the following trigonometric identities.

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Prove the following trigonometric identities.

sec A (1 − sin A) (sec A + tan A) = 1


Prove the following trigonometric identities

tan2 A + cot2 A = sec2 A cosec2 A − 2


Prove the following trigonometric identities.

`(1 + cot A + tan A)(sin A - cos A) = sec A/(cosec^2 A) - (cosec A)/sec^2 A = sin A tan A - cos A cot A`


Prove the following identities:

`cotA/(1 - tanA) + tanA/(1 - cotA) = 1 + tanA + cotA`


What is the value of (1 + cot2 θ) sin2 θ?


(cosec θ − sin θ) (sec θ − cos θ) (tan θ + cot θ) is equal to


Prove the following identity : 

`(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA) = 2`


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`(cot^2θ(secθ - 1))/((1 + sinθ)) = sec^2θ((1-sinθ)/(1 + secθ))`


Prove the following identities:
`1/(sin θ + cos θ) + 1/(sin θ - cos θ) = (2sin θ)/(1 - 2 cos^2 θ)`.


Prove that: `(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(sin^2 A - cos^2 A)`.


Prove that : `tan"A"/(1 - cot"A") + cot"A"/(1 - tan"A") = sec"A".cosec"A" + 1`.


Prove that `(sin^2θ)/(cos θ) + cos θ = sec θ`.


tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.

Activity:

L.H.S. = `square`

= `square (1 - (sin^2θ)/(tan^2θ))`

= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`

= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`

= `tan^2θ (1 - square)`

= `tan^2θ xx square`   ...[1 – cos2θ = sin2θ]

= R.H.S.


Show that tan 7° × tan 23° × tan 60° × tan 67° × tan 83° = `sqrt(3)`.


Prove the following:

`1 + (cot^2 alpha)/(1 + "cosec"  alpha)` = cosec α


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.


If cot θ = `40/9`, find the values of cosec θ and sinθ,

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1 + `square` = cosec2θ

1 + `square` = cosec2θ

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`square/square` = cosec2θ  ......[Taking root on the both side]

cosec θ = `41/9`

and sin θ = `1/("cosec"  θ)`

sin θ = `1/square`

∴ sin θ =  `9/41`

The value is cosec θ = `41/9`, and sin θ = `9/41`


(1 + sin A)(1 – sin A) is equal to ______.


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