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प्रश्न
Prove that:
`sqrt(sec^2A + cosec^2A) = tanA + cotA`
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उत्तर
L.H.S. = `sqrt(sec^2A + cosec^2A)`
= `sqrt(1/cos^2A + 1/sin^2A)`
= `sqrt((sin^2A + cos^2A)/(sin^2Acos^2A)`
= `sqrt(1/(sin^2Acos^2A)`
= `sqrt(1/(sinAcosA))`
R.H.S. = tan A + cot A
= `sinA/cosA + cosA/sinA`
= `(sin^2A + cos^2A)/(sinAcosA)`
= `1/(sinAcosA)`
L.H.S. = R.H.S.
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1 + `square` = cosec2θ
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Find the value of sin2θ + cos2θ
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`
