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प्रश्न
Prove the following identity :
`sinθ(1 + tanθ) + cosθ(1 +cotθ) = secθ + cosecθ`
उत्तर
`sinθ(1 + tanθ) + cosθ(1 +cotθ) = secθ + cosecθ`
LHS = `sinθ(1 + tanθ) + cosθ(1 + cotθ)`
= `sinθ (1 + sinθ/cosθ) + cosθ (1 + cosθ /sinθ)`
= `sinθ((cosθ + sinθ)/cosθ) + cosθ((sinθ + cosθ)/sinθ)`
= `cosθ + sinθ(sinθ /cosθ + cosθ /sinθ)`
= `cosθ + sinθ (1/sinθ 1/cosθ) = secθ + cosecθ `
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संबंधित प्रश्न
Prove the following trigonometric identities.
sec6θ = tan6θ + 3 tan2θ sec2θ + 1
`1+((tan^2 theta) cot theta)/(cosec^2 theta) = tan theta`
cosec4θ − cosec2θ = cot4θ + cot2θ
`sqrt((1-cos theta)/(1+cos theta)) = (cosec theta - cot theta)`
Without using trigonometric identity , show that :
`tan10^circ tan20^circ tan30^circ tan70^circ tan80^circ = 1/sqrt(3)`
Prove that `( 1 + sin θ)/(1 - sin θ) = 1 + 2 tan θ/cos θ + 2 tan^2 θ` .
Prove that cot θ. tan (90° - θ) - sec (90° - θ). cosec θ + 1 = 0.
Choose the correct alternative:
cos θ. sec θ = ?
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 `sqrt((1 + cos "A")/(1 - cos"A"))` = cosec A + cot A
