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If `Cos B = 3/5 and (A + B) =- 90° ,`Find the Value of Sin A. - Mathematics

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प्रश्न

If `cos B = 3/5 and (A + B) =- 90° ,`find the value of sin A.

उत्तर

We have ,

 cos B = `3/5`

  ⇒ ` cos ( 90° - A ) = 3/5       ( As , A+ B = 90°)`

  ∴ sin A = `3/5`

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अध्याय 8: Trigonometric Identities - Exercises 3

APPEARS IN

आरएस अग्रवाल Mathematics [English] Class 10
अध्याय 8 Trigonometric Identities
Exercises 3 | Q 25

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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if x = a cos^3 theta, y = b sin^3 theta` " prove that " `(x/a)^(2/3) + (y/b)^(2/3) = 1`


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`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`


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`1/(1 + cosA) + 1/(1 - cosA) = 2cosec^2A`


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`cot^2A((secA - 1)/(1 + sinA)) + sec^2A((sinA - 1)/(1 + secA)) = 0`


`sqrt((1+cos theta)/(1-cos theta)) + sqrt((1-cos theta )/(1+ cos theta )) = 2 cosec theta`

 


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\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to 

 

 


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`sec^2A.cosec^2A = tan^2A + cot^2A + 2`


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`(1 + cotA)^2 + (1 - cotA)^2 = 2cosec^2A`


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Solution:

In Δ ABC, ∠ABC = 90°, ∠C = θ°

AB2 + BC2 = `square`   .....(Pythagoras theorem)

Divide both sides by AC2

`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`

∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`

But `"AB"/"AC" = square and "BC"/"AC" = square`

∴ `sin^2 theta  + cos^2 theta = square` 


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