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प्रश्न
Prove that:
`"tan A"/(1 + "tan"^2 "A")^2 + "Cot A"/(1 + "Cot"^2 "A")^2 = "sin A cos A"`.
उत्तर
`"tan A"/(1 + "tan"^2 "A")^2 + "cot A"/(1 + "cot"^2 "A")^2 = "sin A cos A"`.
LHS = `"tan A"/(1 + "tan"^2 "A")^2 + "cot A"/(1 + "cot"^2 "A")^2`
LHS = `"tan A"/("sec"^2 "A")^2 + "cot A"/("cosec"^2 "A")^2 ...{( 1 + "tan"^2θ = "sec"^2θ),(1 + "cot"^2θ = "cosec"^2θ):}`
LHS = `"tan A" × 1/("sec"^2 "A")^2 + "cot A" × 1/("cosec"^2 "A")^2`
LHS = `"sin A"/"cos A" × "cos"^4 "A" + "cos A"/"sin A" × "sin"^4 "A" ...{(cosθ = 1/sec θ),(sin θ = 1/"cosecθ"):}`
LHS = sinA cos3A + cosA sin3A
LHS = sinA cosA (cos2A + sin2A)
LHS = sinA cosA.(1) ...(cos2A + sin2A = 1)
LHS = sinA cosA
RHS = sinA cosA
LHS = RHS
Hence proved.
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संबंधित प्रश्न
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Prove the following trigonometric identities.
(sec2 θ − 1) (cosec2 θ − 1) = 1
Prove the following trigonometric identities.
`sin theta/(1 - cos theta) = cosec theta + cot theta`
Prove the following trigonometric identity.
`(sin theta - cos theta + 1)/(sin theta + cos theta - 1) = 1/(sec theta - tan theta)`
Prove the following trigonometric identities.
`1 + cot^2 theta/(1 + cosec theta) = cosec theta`
Prove the following identities:
`(secA - tanA)/(secA + tanA) = 1 - 2secAtanA + 2tan^2A`
If sec A + tan A = p, show that:
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Write the value of `sin theta cos ( 90° - theta )+ cos theta sin ( 90° - theta )`.
What is the value of \[\sin^2 \theta + \frac{1}{1 + \tan^2 \theta}\]
Write True' or False' and justify your answer the following :
The value of the expression \[\sin {80}^° - \cos {80}^°\]
Prove that `sin(90^circ - A).cos(90^circ - A) = tanA/(1 + tan^2A)`
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Prove that: `1/(cosec"A" - cot"A") - 1/sin"A" = 1/sin"A" - 1/(cosec"A" + cot"A")`
Prove the following identities.
`sqrt((1 + sin theta)/(1 - sin theta)` = sec θ + tan θ
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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
