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प्रश्न
If sin θ + cos θ = a and sec θ + cosec θ = b , then the value of b(a2 – 1) is equal to
पर्याय
2a
3a
0
2ab
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उत्तर
2a
Explanation;
Hint:
b(a2 – 1) = (sec θ + cosec θ) [(sin θ + cos θ)2 – 1]
= `1/ cos theta + 1/sin theta` [sin2 θ + cos2 θ + 2 sin θ cos θ – 1]
= `[(sin theta + cos theta)/(sin theta cos theta)]` [1 + 2 sin θ cos θ – 1]
= `[(sin theta + cos theta)/(sin theta cos theta)] xx 2 sin theta cos theta`
= 2(sin θ + cos θ)
= 2a
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संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 + cos A)/sin^2 A = 1/(1 - cos A)`
Prove the following trigonometric identities
tan2 A + cot2 A = sec2 A cosec2 A − 2
`(sec^2 theta-1) cot ^2 theta=1`
`(1+tan^2theta)(1+cot^2 theta)=1/((sin^2 theta- sin^4theta))`
If `(cosec theta - sin theta )= a^3 and (sec theta - cos theta ) = b^3 , " prove that " a^2 b^2 ( a^2+ b^2 ) =1`
If `cos theta = 2/3 , "write the value of" ((sec theta -1))/((sec theta +1))`
Prove that `"cosec" θ xx sqrt(1 - cos^2theta)` = 1
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
Prove that
sec2A – cosec2A = `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")`
Factorize: sin3θ + cos3θ
Hence, prove the following identity:
`(sin^3θ + cos^3θ)/(sin θ + cos θ) + sin θ cos θ = 1`
