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प्रश्न
Prove that:
(cosec θ - sinθ )(secθ - cosθ ) ( tanθ +cot θ) =1
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उत्तर
Taking LHS
(cosec θ - sinθ )(secθ - cos θ ) ( tanθ +cot θ)
`(1/(sin theta )- sin theta )(1/(cos θ )- cosθ )((sin θ)/(cos θ) +(cos θ)/(sin θ))`
`=((1-sin^2 θ)/(sin θ)) ((1- cos ^2θ)/(cos θ)) ((sin^2 θ + cos^2 θ)/(sin θ . cos θ))`
`= (cos^2 θ)/( sin θ) xx (sin^2 θ)/(cos θ ) xx 1/(sinθ . cos θ )` = 1 = RHS
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संबंधित प्रश्न
Prove the following identities:
`(1 - sinA)/(1 + sinA) = (secA - tanA)^2`
` tan^2 theta - 1/( cos^2 theta )=-1`
`1+((tan^2 theta) cot theta)/(cosec^2 theta) = tan theta`
prove that `1/(1 + cos(90^circ - A)) + 1/(1 - cos(90^circ - A)) = 2cosec^2(90^circ - A)`
Without using trigonometric identity , show that :
`sin(50^circ + θ) - cos(40^circ - θ) = 0`
Prove that `tan^3 θ/( 1 + tan^2 θ) + cot^3 θ/(1 + cot^2 θ) = sec θ. cosec θ - 2 sin θ cos θ.`
Without using the trigonometric table, prove that
cos 1°cos 2°cos 3° ....cos 180° = 0.
Prove that: `1/(cosec"A" - cot"A") - 1/sin"A" = 1/sin"A" - 1/(cosec"A" + cot"A")`
`5/(sin^2θ) - 5cot^2θ`, complete the activity given below.
Activity:
`5/(sin^2θ) - 5cot^2θ`
= `square (1/(sin^2θ) - cot^2θ)`
= `5(square - cot^2θ) ...[1/(sin^2θ) = square]`
= 5(1)
= `square`
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
