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प्रश्न
If `sec theta + tan theta = p,` prove that
(i)`sec theta = 1/2 ( p+1/p) (ii) tan theta = 1/2 ( p- 1/p) (iii) sin theta = (p^2 -1)/(p^2+1)`
उत्तर
(i) We have , `sec theta + tan theta = p` ....................(1)
`⇒ (sec theta + tan theta )/1 xx (sec theta - tan theta )/( sec theta - tan theta ) = p`
`⇒ (sec ^2 theta - tan^2 theta )/( sec theta - tan theta) = p`
`⇒ 1/ (sec theta - tan theta ) =p`
`⇒ sec theta - tan theta = 1/ p` .........................(2)
Adding (1) and (2) , We get
2` sec theta = p + 1/p`
`⇒ sec theta = 1/2 ( p+1/p)`
(ii) subtracting (2) feom (1) , We get
`2 tan theta = (p - 1/p)`
`⇒ tan theta = 1/2 ( p-1/p)`
(iii) Using (i) and (ii) , We get
`sin theta = tantheta/ sec theta`
=`(1/2(p-1/p))/(1/2 (p+1/p)`
=`(((p^2-1)/p))/(((p^2+1))/p)`
∴ `sin theta = (p^2-1)/(p^2 +1)`
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sec2θ = 1 + `square` ......[Fundamental trigonometric identity]
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In Δ ABC, ∠ABC = 90°, ∠C = θ°
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