Ek admi ke paas 9 diamond the 8 diamond 8 gm ke the aur ek 9 gm ka galti se uske sare diamond mix ho jate hai to wehe ek sunar ke paas jata hai diamond ko tulwane pqr sunaar usko bolta hai ki mai tumko sirf do baar isko tolne dunga to batao wo 9 gm ka diamond kaise dhund payega ?
Answers
Answer:
Solution−
Consider,
\begin{gathered}\sf \: \dfrac{1+cos \theta + sin \theta}{1+ cos \theta - sin \theta} \\ \\ \end{gathered}
1+cosθ−sinθ
1+cosθ+sinθ
Divide numerator and denominator by cos\thetaθ , we get
\begin{gathered}\sf \: = \: \dfrac{ \dfrac{1+cos \theta + sin \theta}{cos\theta }}{ \dfrac{1+ cos \theta - sin \theta}{cos\theta }} \\ \\ \end{gathered}
=
cosθ
1+cosθ−sinθ
cosθ
1+cosθ+sinθ
\begin{gathered}\sf \: = \: \dfrac{\dfrac{1}{cos\theta } + \dfrac{cos\theta }{cos\theta } + \dfrac{sin\theta }{cos\theta } }{\dfrac{1}{cos\theta } + \dfrac{cos\theta }{cos\theta } - \dfrac{sin\theta }{cos\theta }} \\ \\ \end{gathered}
=
cosθ
1
+
cosθ
cosθ
−
cosθ
sinθ
cosθ
1
+
cosθ
cosθ
+
cosθ
sinθ
\begin{gathered}\sf \: = \: \dfrac{sec\theta + 1 + tan\theta }{sec\theta + 1 - tan\theta } \\ \\ \end{gathered}
=
secθ+1−tanθ
secθ+1+tanθ
\begin{gathered}\boxed{ \sf{ \: \because \: \frac{1}{cosx} = secx \: \: and \: \: \frac{sinx}{cosx} = tanx \: }} \\ \\ \end{gathered}
∵
cosx
1
=secxand
cosx
sinx
=tanx
\begin{gathered}\sf \: = \: \dfrac{(sec\theta + tan\theta) + 1 }{sec\theta + 1 - tan\theta } \\ \\ \end{gathered}
=
secθ+1−tanθ
(secθ+tanθ)+1
\begin{gathered}\sf \: = \: \dfrac{(sec\theta + tan\theta) + {sec}^{2}\theta - {tan}^{2} \theta }{sec\theta + 1 - tan\theta } \\ \\ \end{gathered}
=
secθ+1−tanθ
(secθ+tanθ)+sec
2
θ−tan
2
θ
\begin{gathered}\boxed{ \sf{ \: \because \: {sec}^{2}x - {tan}^{2}x = 1 \: }} \\ \\ \end{gathered}
∵sec
2
x−tan
2
x=1
\begin{gathered}\sf \: = \: \dfrac{(sec\theta + tan\theta) + (sec\theta + tan\theta )(sec\theta - tan\theta )}{sec\theta + 1 - tan\theta } \\ \\ \end{gathered}
=
secθ+1−tanθ
(secθ+tanθ)+(secθ+tanθ)(secθ−tanθ)
\begin{gathered}\boxed{ \sf{ \: \because \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: }} \\ \\ \end{gathered}
∵x
2
−y
2
=(x+y)(x−y)
\begin{gathered}\sf \: = \: \dfrac{(sec\theta + tan\theta)[1 + sec\theta - tan\theta ]}{sec\theta + 1 - tan\theta } \\ \\ \end{gathered}
=
secθ+1−tanθ
(secθ+tanθ)[1+secθ−tanθ]
\begin{gathered}\sf \: = \: sec\theta + tan\theta \\ \\ \end{gathered}
=secθ+tanθ
\begin{gathered}\sf \: = \: \dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \\ \end{gathered}
=
cosθ
1
+
cosθ
sinθ
\begin{gathered}\boxed{ \sf{ \: \because \: \frac{1}{cosx} = secx \: \: and \: \: \frac{sinx}{cosx} = tanx \: }} \\ \\ \end{gathered}
∵
cosx
1
=secxand
cosx
sinx
=tanx
\begin{gathered}\sf \: = \: \dfrac{1 + sin\theta }{cos\theta } \\ \\ \end{gathered}
=
cosθ
1+sinθ
Hence,
\begin{gathered}\bf\implies \:\dfrac{1}{cos\theta } + \dfrac{cos\theta }{cos\theta } + \dfrac{sin\theta }{cos\theta } = \: \dfrac{1 + sin\theta }{cos\theta } \\ \\ \\ \end{gathered}
⟹
cosθ
1
+
cosθ
cosθ
+
cosθ
sinθ
=
cosθ
1+sinθ