BHAI KONSE CLASS MEIN HO APP LOG
Answers
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