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प्रश्न
` (sin theta + cos theta )/(sin theta - cos theta ) + ( sin theta - cos theta )/( sin theta + cos theta) = 2/ ((1- 2 cos^2 theta))`
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उत्तर
LHS = `(sin theta + cos theta )/(sin theta - cos theta ) + ( sin theta - cos theta )/( sin theta + cos theta)`
=` ((sin theta + cos theta )^2 + ( sin theta - cos theta)^2)/(( sin theta - cos theta ) ( sin theta + cos theta))`
=`( sin^2 theta + cos^2 theta + 2 sin theta cos theta + sin^2 theta + cos^2 theta - 2 sin theta cos theta)/((sin^2 theta - cos^2 theta))`
=`(1+1)/((- cos^ 2theta )- cos^2 theta) (∵ sin^ 2theta + cos^2 theta =1)`
=`2/(1-2 cos^2 theta)`
= RHS
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संबंधित प्रश्न
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x = 3 cosec θ + 4 cot θ
y = 4 cosec θ – 3 cot θ
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Write True' or False' and justify your answer the following :
The value of sin θ+cos θ is always greater than 1 .
Prove the following identity :
`tanA - cotA = (1 - 2cos^2A)/(sinAcosA)`
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`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
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a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p2 – q2 is equal to
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If tan θ = `7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ......[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` .......`[cos theta = 1/sectheta]`
Prove that `(1 + sec "A")/"sec A" = (sin^2"A")/(1 - cos"A")`
