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If `sec theta + tan theta = x," find the value of " sec theta`
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We have ,
`sec theta + tan theta = x ............(i)`
⇒ `(sec theta + tan theta )/1 xx (sec theta- tan theta )/(sec theta - tan theta ) = x`
`⇒ (sec ^2 theta - tan^2 theta )/( sec theta - tan theta) = x`
`⇒1/ (sec theta - tan theta ) = x/1`
`⇒ sec theta - tan theta = 1/x ` ............(ii)
ЁЭР┤ЁЭССЁЭССЁЭСЦЁЭСЫЁЭСФ (ЁЭСЦ)ЁЭСОЁЭСЫЁЭСС (ЁЭСЦЁЭСЦ), ЁЭСдЁЭСТ ЁЭСФЁЭСТЁЭСб
`2 sec theta = x+ 1/x`
⇒` 2 sec theta = (x^2+1)/x`
∴ `sec theta = (x^2 +1)/(2x)`
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Prove that cos2θ . (1 + tan2θ) = 1. Complete the activity given below.
Activity:
L.H.S = `square`
= `cos^2theta xx square .....[1 + tan^2theta = square]`
= `(cos theta xx square)^2`
= 12
= 1
= R.H.S
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Prove the following:
`sintheta/(1 + cos theta) + (1 + cos theta)/sintheta` = 2cosecθ
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
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