Question

Nhi yaar, 1000 answer ka itazar kr rha hun ruk ja thoda​

Answer 1

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