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प्रश्न
Prove the following identity :
`sinθ(1 + tanθ) + cosθ(1 +cotθ) = secθ + cosecθ`
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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 `
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`sqrt((1 - cos theta)/(1 + cos theta)) = cosec theta - cot theta`
Prove the following identities:
`(1 - cosA)/sinA + sinA/(1 - cosA)= 2cosecA`
`1/((1+tan^2 theta)) + 1/((1+ tan^2 theta))`
If `x/(a cosθ) = y/(b sinθ) "and" (ax)/cosθ - (by)/sinθ = a^2 - b^2 , "prove that" x^2/a^2 + y^2/b^2 = 1`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`tan35^circ cot(90^circ - θ) = 1`
Prove the following identities.
cot θ + tan θ = sec θ cosec θ
If x = a tan θ and y = b sec θ then
If tan θ × A = sin θ, then A = ?
If sec θ + tan θ = `sqrt(3)`, complete the activity to find the value of sec θ – tan θ
Activity:
`square` = 1 + tan2θ ......[Fundamental trigonometric identity]
`square` – tan2θ = 1
(sec θ + tan θ) . (sec θ – tan θ) = `square`
`sqrt(3)*(sectheta - tan theta)` = 1
(sec θ – tan θ) = `square`
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
