Question

Dimaag lagao

A Big Challenge 4 U!!
Time 1 Day
Total Marks 100

Ek Dada
Ek Baap
Ek Pota

3no Ki Total Age 140 Years Hai
Pota jitne Month Ka Dada utne Saal ka
Pota jinte din ka Baap utne week ka hai

Batao 3no Kitne kitne Saal k Hain?
Reply if u r Genius

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