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प्रश्न
Show that: `tan "A"/(1 + tan^2 "A")^2 + cot "A"/(1 + cot^2 "A")^2 = sin"A" xx cos"A"`
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उत्तर
Proof: L.H.S. = `tan"A"/(1 + tan^2 "A")^2 + cot"A"/(1 + cot^2 "A")^2`
= `tan "A"/(sec^2"A")^2 + cot "A"/("cosec"^2"A")^2` ......`[(∵ 1 + cot^2θ = "cosec"^2θ),(1 + tan^2θ = sec^2θ)]`
= `tan "A"/sec^4"A" + cot "A"/("cosec"^4"A")`
= `sin "A"/cos "A" xx 1/(sec^4 "A") + cos "A"/sin "A" xx 1/("cosec"^4 "A")`
= `sin "A"/cos "A" xx cos^4"A" + cos "A"/sin "A" xx sin^4"A"`
= sinA × cos3A + cosA × sin3A
= sinA cosA (cos2A + sin2A)
= sinA cosA (1) ......[∵ cos2A + sin2A = 1]
= sinA.cosA
= R.H.S
L.H.S. = R.H.S.
Hence proved.
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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.
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(sin α + cos α)(tan α + cot α) = sec α + cosec α
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Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
