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प्रश्न
In the figure of ΔPQR , ∠P = θ° and ∠R =∅° find
(i) `sqrt(X +1) cot ∅`
(ii)`sqrt( x^3 + x ^2) tantheta`
(iii) cos θ

योग
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उत्तर

In Δ𝑃𝑄𝑅, ∠𝑄 = 900,
Using Pythagoras theorem, we get
𝑃𝑄 = `sqrt(PR^2 − QR^2)`
= `sqrt ((x + 2)^2 − x^2)`
= `sqrt(x^2 + 4x + 4 − x^2)`
= `sqrt(4 (x+ 1))`
= `2sqrt(x + 1)`
Now,
(i) `(sqrt(x+1) cot theta`
=`(sqrt(x+1))xx(QR)/(PQ)`
=`(sqrt(x+1))xx x/(2 sqrt(x+1))`
=`x/2`
(ii) `(sqrt(x^3+x^2)) tan theta`
= `(sqrt(x^2(x+1)))xx(QR)/(PQ)`
`=x sqrt((x+1))xx x/(2sqrt(x+1)`
= `x^2/2`
(iii) cos θ
=`(PQ)/(PR) theta=(2sqrt(x+1))/(x+2)`
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