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प्रश्न
Prove the following identities:
`1/(secA + tanA) = secA - tanA`
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उत्तर १
L.H.S. = `1/(secA + tanA)`
= `1/(1/cosA + sinA/cosA)`
= `1/((1 + sinA)/cosA)`
= `cosA/(1 + sinA) xx (1 - sinA)/(1 + sinA)`
= `(cosA(1 - sinA))/((1)^2 - sin^2A)`
= `(cosA(1 - sinA))/cos^2A`
= `1/cosA - sinA/cosA`
= sec A – tan A
L.H.S. = R.H.S.
Hence proved.
उत्तर २
L.H.S = `1/(secA + tanA)`
= `((secA - tanA))/((secA + tanA)(secA - tanA))` ...((Multiply Num. and Deno. by sec A – tan A)
= `(secA - tanA)/(sec^2A - tan^2A)`
= `(secA - tanA)/1` ...[∵ sec2 A – tan2 A = 1]
= sec A – tan A
= R.H.S.
संबंधित प्रश्न
Prove the following identities:
`(i) 2 (sin^6 θ + cos^6 θ) –3(sin^4 θ + cos^4 θ) + 1 = 0`
`(ii) (sin^8 θ – cos^8 θ) = (sin^2 θ – cos^2 θ) (1 – 2sin^2 θ cos^2 θ)`
Prove the following trigonometric identities.
`(1 + cos A)/sin^2 A = 1/(1 - cos A)`
Prove that: `sqrt((sec theta - 1)/(sec theta + 1)) + sqrt((sec theta + 1)/(sec theta - 1)) = 2 cosec theta`
`(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))=1`
Write the value of `4 tan^2 theta - 4/ cos^2 theta`
If `cos theta = 2/3 , "write the value of" ((sec theta -1))/((sec theta +1))`
If \[\sin \theta = \frac{1}{3}\] then find the value of 2cot2 θ + 2.
Prove that: `(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(sin^2 A - cos^2 A)`.
Show that `(cos^2(45^circ + θ) + cos^2(45^circ - θ))/(tan(60^circ + θ) tan(30^circ - θ)) = 1`.
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`
