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प्रश्न
If cosθ + sinθ = `sqrt2` cosθ, show that cosθ - sinθ = `sqrt2` sinθ.
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उत्तर १
(cosθ + sinθ)2 = (`sqrt2`. cosθ)2
cos2θ + sin2θ + 2.cosθ.sinθ = 2cos2θ
1 + 2.cosθ.sinθ = 2cos2θ
2.cosθ.sinθ = 2cos2θ − 1
(cosθ.sinθ)2 = cos2θ + sin2θ − 2.cosθ.sinθ
= 1 − (2.cos2θ − 1)
= 1 − 2.cos2θ +1
= 2 − 2.cos2θ
= 2(1 − cos2θ)
cosθ − sinθ = `sqrt(2sin^2θ)`
= `sqrt2`sinθ
Hence proved
उत्तर २
संबंधित प्रश्न
Prove the following identities:
`sqrt((1 + sinA)/(1 - sinA)) = cosA/(1 - sinA)`
`sin theta / ((1+costheta))+((1+costheta))/sin theta=2cosectheta`
If tan A =` 5/12` , find the value of (sin A+ cos A) sec A.
9 sec2 A − 9 tan2 A is equal to
Prove the following identity :
`sqrt((secq - 1)/(secq + 1)) + sqrt((secq + 1)/(secq - 1))` = 2 cosesq
Find the value of `θ(0^circ < θ < 90^circ)` if :
`cos 63^circ sec(90^circ - θ) = 1`
Prove that `( tan A + sec A - 1)/(tan A - sec A + 1) = (1 + sin A)/cos A`.
Prove that: `1/(cosec"A" - cot"A") - 1/sin"A" = 1/sin"A" - 1/(cosec"A" + cot"A")`
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
Prove the following:
`tanA/(1 + sec A) - tanA/(1 - sec A)` = 2cosec A
