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(sec θ + tan θ) . (sec θ – tan θ) = ? - Geometry Mathematics 2

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प्रश्न

(sec θ + tan θ) . (sec θ – tan θ) = ?

योग
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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)]`

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अध्याय 6: Trigonometry - Q.1 (B)

संबंधित प्रश्न

Prove the following identities:

`(i) cos4^4 A – cos^2 A = sin^4 A – sin^2 A`

`(ii) cot^4 A – 1 = cosec^4 A – 2cosec^2 A`

`(iii) sin^6 A + cos^6 A = 1 – 3sin^2 A cos^2 A.`


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`1/(sec A - 1) + 1/(sec A + 1) = 2 cosec A cot A`


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`[tan θ + 1/cos θ]^2 + [tan θ - 1/cos θ]^2 = 2((1 + sin^2 θ)/(1 - sin^2 θ))`


Prove the following trigonometric identities

If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2 − y2 = a2 − b2


Prove the following identities:

`(1 + sin A)/(1 - sin A) = (cosec  A + 1)/(cosec  A - 1)`


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`sqrt((1 - sinA)/(1 + sinA)) = cosA/(1 + sinA)`


Show that : `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec A cosec A`


If x = a cos θ and y = b cot θ, show that:

`a^2/x^2 - b^2/y^2 = 1` 


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Prove the following identity :

`sinθ(1 + tanθ) + cosθ(1 +cotθ) = secθ + cosecθ` 


Prove the following identity :

secA(1 - sinA)(secA + tanA) = 1


Prove the following identity : 

`sin^4A + cos^4A = 1 - 2sin^2Acos^2A`


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tan (55° + x) = cot (35° – x)


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The value of the expression [cosec(75° + θ) – sec(15° – θ) – tan(55° + θ) + cot(35° – θ)] is ______.


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

We have, 1 + cot2θ = cosec2θ

1 + `square` = cosec2θ

1 + `square` = cosec2θ

`(square + square)/square` = cosec2θ

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

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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`


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