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प्रश्न
Prove that `(sin (90° - θ))/cos θ + (tan (90° - θ))/cot θ + (cosec (90° - θ))/sec θ = 3`.
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उत्तर
LHS = `(sin (90° - θ))/cos θ + (tan (90° - θ))/cot θ + (cosec (90° - θ))/sec θ `
= `cos θ/cos θ + cot θ/cot θ + sec θ/sec θ`
= 1 + 1 + 1
= 3
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
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Prove the following identities:
(1 + tan A + sec A) (1 + cot A – cosec A) = 2
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If `(cot theta ) = m and ( sec theta - cos theta) = n " prove that " (m^2 n)(2/3) - (mn^2)(2/3)=1`
If`( 2 sin theta + 3 cos theta) =2 , " prove that " (3 sin theta - 2 cos theta) = +- 3.`
Prove the following identities:
`(tan"A"+tan"B")/(cot"A"+cot"B")=tan"A"tan"B"`
Prove that (cosec A - sin A)( sec A - cos A) sec2 A = tan A.
Prove that cosec θ – cot θ = `sin theta/(1 + cos theta)`
Statement 1: sin2θ + cos2θ = 1
Statement 2: cosec2θ + cot2θ = 1
Which of the following is valid?
