Question

$\frac{1 + \frac{1 }{ \cos(x) } }{ \frac{1}{ \cos(x) } }$
simplify it

​

Answer 1

Answer:

Here's ur answer mate:-

As per the information provided in the question, We have :

\begin{gathered} \longmapsto \rm \dfrac{1 + \dfrac{1 }{ \cos(x) } }{ \dfrac{1}{ \cos(x) } }\\\\\end{gathered}

⟼

cos(x)

1

1+

cos(x)

1

We are asked to simplify it.

Using the fraction rule i.e a/b/c = a × c/b, Thus,

\begin{gathered} \\\longmapsto \rm \dfrac{1 + \dfrac{1 }{ \cos(x) } }{ \dfrac{1}{ \cos(x) } }\\\\\end{gathered}

⟼

cos(x)

1

1+

cos(x)

1

\begin{gathered}\longmapsto \rm 1 + \frac{1}{ \cos(x) } \times \frac{ \cos(x) }{1} \\\\\end{gathered}

⟼1+

cos(x)

1

×

1

cos(x)

\begin{gathered}\longmapsto \rm \bigg(1 + \frac{1}{ \cos(x) } \bigg) { \cos(x) } \\\\\end{gathered}

⟼(1+

cos(x)

1

)cos(x)

\begin{gathered}\longmapsto \rm (1 + \sec(x) ){ \cos(x) } \\\\\end{gathered}

⟼(1+sec(x))cos(x)

\begin{gathered}\longmapsto \rm 1 + \cos(x) \\\\\end{gathered}

⟼1+cos(x)

This can't be further simplified. Hence, It 1 + cos x is the required answer.

Learn more!

\begin{gathered}\begin{gathered}\qquad \sf \: ( I ) \:sin^2\:\theta \:+\:cos^2 \:\:=\:1\:\\\\\end{gathered}\end{gathered}

(I)sin

2

θ+cos

2

=1

\begin{gathered}\begin{gathered}\qquad \sf \: ( II ) \:sin^2\:\theta \:=\: 1 \:-\:cos^2 \:\:\:\\\\\end{gathered}\end{gathered}

(II)sin

2

θ=1−cos

2

\begin{gathered}\begin{gathered}\qquad \sf \: ( III ) \:\:cos^2 \:\:=\:1 \:-\:sin^2\:\theta \:\\\\\end{gathered}\end{gathered}

(III)cos

2

=1−sin

2

θ

\begin{gathered}\begin{gathered}\qquad \sf \: ( IV ) \:\:\:1 \:+\:cot^2\:\theta \:=\: cosec^2\:\theta \:\\\\\end{gathered}\end{gathered}

(IV)1+cot

2

θ=cosec

2

θ

\begin{gathered}\begin{gathered}\qquad \sf \: ( V ) \:\:\:cosec^2\:\theta \:-\:cot^2\:\theta \:=\: 1 \:\\\\\end{gathered}\end{gathered}

(V)cosec

2

θ−cot

2

θ=1

\begin{gathered}\begin{gathered}\qquad \sf \: ( VI ) \:\:\:cosec^2\:\theta \:=\:cot^2\:\theta \:+\: 1 \:\\\\\end{gathered}\end{gathered}

(VI)cosec

2

θ=cot

2

θ+1

Hope it helps uh!