UG Element Math Solutions Part 3

Elements of Mathematics Solutions for Class 11th ( Part – 3 ) – Unbox Goodies

Class 11thElements of Math Solutions

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Solutions Start From Here ( Part – 3 )

(i)\text{sine}\theta (written as \text{sin}\theta ) =\frac{{\text{ }\!\!~\!\!\text{ perpendicular }\!\!~\!\!\text{ }}}{{\text{ }\!\!~\!\!\text{ hypotenuse }\!\!~\!\!\text{ }}}=\frac{y}{r}


(ii) \text{cosine}\theta ( written as \text{cos}\theta )=\frac{{\text{ }\!\!~\!\!\text{ base }\!\!~\!\!\text{ }}}{{\text{ }\!\!~\!\!\text{ hypotenuse }\!\!~\!\!\text{ }}}=\frac{x}{r}


(iii) tangent \theta (written as \text{tan}\theta ) =\frac{{\text{ }\!\!~\!\!\text{ perpendicular }\!\!~\!\!\text{ }}}{{\text{ }\!\!~\!\!\text{ base }\!\!~\!\!\text{ }}}=\frac{y}{x}


(iv) \text{cotangent}\theta (written as \text{cot}\theta )=\frac{{\text{ }\!\!~\!\!\text{ base }\!\!~\!\!\text{ }}}{{\text{ }\!\!~\!\!\text{ perpendicular }\!\!~\!\!\text{ }}}=\frac{x}{y}


(v) \text{secant}\theta ( written as \text{sec}\theta )=\frac{{\text{ }\!\!~\!\!\text{ hypotenuse }\!\!~\!\!\text{ }}}{{\text{ }\!\!~\!\!\text{ base }\!\!~\!\!\text{ }}}=\frac{r}{x}


(vi) \text{cosecant}\theta (written as \text{cosec}\theta )=\frac{{\text{ }\!\!~\!\!\text{ hypotenuse }\!\!~\!\!\text{ }}}{{\text{ }\!\!~\!\!\text{ perpendicular }\!\!~\!\!\text{ }}}=\frac{r}{y}


Here, we also recall some important identities of trigonometric ratios.


(1) \text{tan}\theta =\frac{{\text{sin}\theta }}{{\text{cos}\theta }}


(2) \text{cot}\theta =\frac{{\text{cos}\theta }}{{\text{sin}\theta }}


(3) \text{cosec}\theta =\frac{1}{{\text{sin}\theta }}


(4) \text{sec}\theta =\frac{1}{{\text{cos}\theta }}


(5) \text{cot}\theta =\frac{1}{{\text{tan}\theta }}


(6) \text{si}{{\text{n}}^{2}}\theta +\text{co}{{\text{s}}^{2}}\theta =1 i.e., \text{si}{{\text{n}}^{2}}\theta =1-\text{co}{{\text{s}}^{2}}\theta i.e., \text{co}{{\text{s}}^{2}}\theta =1-\text{si}{{\text{n}}^{2}}\theta .


(7) 1+\text{ta}{{\text{n}}^{2}}\theta =\text{se}{{\text{c}}^{2}}\theta i.e., \text{se}{{\text{c}}^{2}}\theta -\text{ta}{{\text{n}}^{2}}\theta =1 i.e., \text{ta}{{\text{n}}^{2}}\theta =\text{se}{{\text{c}}^{2}}\theta -1


(8) 1+\text{co}{{\text{t}}^{2}}\theta =\text{cose}{{\text{c}}^{2}}\theta i.e., \text{cose}{{\text{c}}^{2}}\theta -\text{co}{{\text{t}}^{2}}\theta =1 i.e., \text{co}{{\text{t}}^{2}}\theta =\text{cose}{{\text{c}}^{2}}\theta -1.

 

Example 1. Prove the following:


(i) \text{si}{{\text{n}}^{4}}\theta -\text{co}{{\text{s}}^{4}}\theta =\text{si}{{\text{n}}^{2}}\theta -\text{co}{{\text{s}}^{2}}\theta


(ii) \text{si}{{\text{n}}^{6}}\theta +\text{co}{{\text{s}}^{6}}\theta =1-3\text{si}{{\text{n}}^{2}}\theta \text{co}{{\text{s}}^{2}}\theta


Solution.

(i) L.H.S.

=\text{si}{{\text{n}}^{4}}\theta -\text{co}{{\text{s}}^{4}}\theta

 

={{\left( {\text{si}{{\text{n}}^{2}}\theta } \right)}^{2}}-{{\left( {\text{co}{{\text{s}}^{2}}\theta } \right)}^{2}}

 

=\left( {\text{si}{{\text{n}}^{2}}\theta +\text{co}{{\text{s}}^{2}}\theta } \right)\left( {\text{si}{{\text{n}}^{2}}\theta -\text{co}{{\text{s}}^{2}}\theta } \right)

 

=\text{si}{{\text{n}}^{2}}\theta -\text{co}{{\text{s}}^{2}}\theta =\text{ }\!\!~\!\!\text{ R}.\text{H}.\text{S}.\text{ }\!\!~\!\!\text{ }\left[ {\because \text{si}{{\text{n}}^{2}}\theta +\text{co}{{\text{s}}^{2}}\theta =1} \right]

 

\text{(ii) }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ L}\text{.H}\text{.S}\text{. }\!\!~\!\!\text{ = }\!\!~\!\!\text{ si}{{\text{n}}^{6}}\theta +\text{co}{{\text{s}}^{6}}\theta

 

=~{{\left( {\text{si}{{\text{n}}^{2}}\theta } \right)}^{3}}+{{\left( {\text{co}{{\text{s}}^{2}}\theta } \right)}^{3}}

 

=~{{\left( {\text{si}{{\text{n}}^{2}}\theta } \right)}^{2}}-{{\left( {\text{co}{{\text{s}}^{2}}\theta } \right)}^{2}}

 

=~\left( {\text{si}{{\text{n}}^{2}}\theta +\text{co}{{\text{s}}^{2}}\theta } \right)\left( {\text{si}{{\text{n}}^{2}}\theta -\text{co}{{\text{s}}^{2}}\theta } \right)

 

={{\left( {\text{si}{{\text{n}}^{2}}\theta +\text{co}{{\text{s}}^{2}}\theta } \right)}^{3}}-3\text{si}{{\text{n}}^{2}}\theta \text{co}{{\text{s}}^{2}}\theta \left( {\text{si}{{\text{n}}^{2}}\theta +\text{co}{{\text{s}}^{2}}\theta } \right)

 

=1-3\text{si}{{\text{n}}^{2}}\theta \text{co}{{\text{s}}^{2}}\theta =\text{ }\!\!~\!\!\text{ R}.\text{H}.\text{S}.\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\left[ {\because \text{si}{{\text{n}}^{2}}\theta +\text{co}{{\text{s}}^{2}}\theta =1} \right]

 

={{\left( {\text{si}{{\text{n}}^{2}}\theta } \right)}^{3}}+{{\left( {\text{co}{{\text{s}}^{2}}\theta } \right)}^{3}}

 

={{\left( {\text{si}{{\text{n}}^{2}}\theta +\text{co}{{\text{s}}^{2}}\theta } \right)}^{3}}-3\text{si}{{\text{n}}^{2}}\theta \text{co}{{\text{s}}^{2}}\theta \left( {\text{si}{{\text{n}}^{2}}\theta +\text{co}{{\text{s}}^{2}}\theta } \right)

 

=1-3\text{si}{{\text{n}}^{2}}\theta \text{co}{{\text{s}}^{2}}\theta = R.H.S. \left[ {\because \text{si}{{\text{n}}^{2}}\theta +\text{co}{{\text{s}}^{2}}\theta =1} \right]

 

Example 2. Prove that :


(i) \text{ta}{{\text{n}}^{2}}\theta -\text{si}{{\text{n}}^{2}}\theta =\text{ta}{{\text{n}}^{2}}\theta \text{si}{{\text{n}}^{2}}\theta


(ii) \frac{{1-\text{ta}{{\text{n}}^{2}}\text{ }\!\!~\!\!\text{ A}}}{{1+\text{ta}{{\text{n}}^{2}}\text{ }\!\!~\!\!\text{ A}}}=\text{co}{{\text{s}}^{2}}\text{ }\!\!~\!\!\text{ A}-\text{si}{{\text{n}}^{2}}\text{ }\!\!~\!\!\text{ A}

 

(iii) \frac{{\text{tan}\theta -\text{cot}\theta }}{{\text{sin}\theta \text{cos}\theta }}=\text{se}{{\text{c}}^{2}}\theta -\text{cose}{{\text{c}}^{2}}\theta


(iv) \frac{{\text{cosec}\theta }}{{\text{cot}\theta +\text{tan}\theta }}=\text{cos}\theta .


Solution. (i) L.H.S. =\text{ta}{{\text{n}}^{2}}\theta -\text{si}{{\text{n}}^{2}}\theta

 

=\frac{{\text{si}{{\text{n}}^{2}}\theta }}{{\text{co}{{\text{s}}^{2}}\theta }}-\text{si}{{\text{n}}^{2}}\theta =\frac{{\text{si}{{\text{n}}^{2}}\theta -\text{si}{{\text{n}}^{2}}\theta \text{co}{{\text{s}}^{2}}\theta }}{{\text{co}{{\text{s}}^{2}}\theta }}

 

=\frac{{\text{si}{{\text{n}}^{\text{*}}}\theta \left( {1-\text{co}{{\text{s}}^{2}}\theta } \right)}}{{\text{co}{{\text{s}}^{2}}\theta }}=\frac{{\text{si}{{\text{n}}^{2}}\theta }}{{\text{co}{{\text{s}}^{2}}\theta }}\cdot \text{si}{{\text{n}}^{2}}\theta

 

\begin{array}{*{20}{r}} {} & {} \\ {} & {} \\ {} & {~=\text{ta}{{\text{n}}^{2}}\theta \text{si}{{\text{n}}^{2}}\theta =\text{ }\!\!~\!\!\text{ R}.\text{H}.\text{S}.\text{ }\!\!~\!\!\text{ }} \end{array}

 

(ii) L.H.S. =\frac{{1-\text{ta}{{\text{n}}^{2}}\text{ }\!\!~\!\!\text{ A}}}{{1+\text{ta}{{\text{n}}^{2}}\text{ }\!\!~\!\!\text{ A}}}

 

=\frac{{1-\frac{{\text{si}{{\text{n}}^{2}}\text{ }\!\!~\!\!\text{ A}}}{{\text{co}{{\text{s}}^{2}}\text{ }\!\!~\!\!\text{ A}}}}}{{1+\frac{{\text{si}{{\text{n}}^{2}}\text{ }\!\!~\!\!\text{ A}}}{{\text{co}{{\text{s}}^{2}}\text{ }\!\!~\!\!\text{ A}}}}}

 

=\frac{{\text{co}{{\text{s}}^{2}}\text{ }\!\!~\!\!\text{ A}-\text{si}{{\text{n}}^{2}}\text{ }\!\!~\!\!\text{ A}}}{{\text{co}{{\text{s}}^{2}}\text{ }\!\!~\!\!\text{ A}+\text{si}{{\text{n}}^{2}}\text{ }\!\!~\!\!\text{ A}}}


=\text{co}{{\text{s}}^{2}}\mathbf{A}-\text{si}{{\text{n}}^{2}}\mathbf{A}= R.H.S.

 

\text{ }\!\!~\!\!\text{ }\left[ {\because \text{co}{{\text{s}}^{2}}A+\text{si}{{\text{n}}^{2}}A=1} \right]

 

(iii) \text{L}\text{.H}\text{.S}\text{. }\!\!~\!\!\text{ }~=\frac{{\text{tan}\theta -\text{cot}\theta }}{{\text{sin}\theta \text{cos}\theta }}

 

=\frac{{\frac{{\text{sin}\theta }}{{\text{cos}\theta }}}}{{\text{sin}\theta \text{cos}\theta }}-\frac{{\frac{{\text{cos}\theta }}{{\text{sin}\theta }}}}{{\text{sin}\theta \text{cos}\theta }}

 

=\frac{1}{{\text{co}{{\text{s}}^{2}}\theta }}-\frac{1}{{\text{si}{{\text{n}}^{2}}\theta }}

 

=\text{se}{{\text{c}}^{2}}\theta -\text{cose}{{\text{c}}^{2}}\theta =\text{ }\!\!~\!\!\text{ R}.\text{H}.\text{S}.

 

(iv) \text{L}\text{.H}\text{.S}\text{.}~=\frac{{\text{cosec}\theta }}{{\text{cot}\theta +\text{tan}\theta }}

 

=\frac{{\frac{1}{{\text{sin}\theta }}}}{{\frac{{\text{cos}\theta }}{{\text{sin}\theta }}+\frac{{\text{sin}\theta }}{{\text{cos}\theta }}}}

 

=\frac{1}{{\frac{{\text{sin}\theta }}{{\text{co}{{\text{s}}^{2}}\theta +\text{si}{{\text{n}}^{2}}\theta }}}}~=\frac{1}{{\text{sin}\theta }}\times \frac{{\text{sin}\theta \text{cos}\theta }}{1}

 

=\text{cos}\theta =\text{ }\!\!~\!\!\text{ R}.\text{H}.\text{S}.

 

Example 3. Prove that:


(i) \sqrt{{\frac{{1+\text{sin}A}}{{1-\text{sin}A}}}}=\text{sec}A+\text{tan}A


(ii) \sqrt{{\frac{{1-\text{cos}\theta }}{{1+\text{cos}\theta }}}}=\text{cosec}\theta -\text{cot}\theta


Solution. (i) L.H.S. =\sqrt{{\frac{{1+\text{sinA}}}{{1-\text{sinA}}}}}

 

=\frac{{\sqrt{{1+\text{sinA}}}}}{{\sqrt{{1-\text{sinA}}}}}\times \frac{{\sqrt{{1+\text{sinA}}}}}{{\sqrt{{1+\text{sinA}}}}}[Rationalism the numerator]

 

=\frac{{1+\text{sinA}}}{{\sqrt{{\left( {1-\text{sinA}} \right)\left( {1+\text{sinA}} \right)}}}}

 

=\frac{{1+\text{sinA}}}{{\sqrt{{1-\text{si}{{\text{n}}^{2}}\text{ }\!\!~\!\!\text{ A}}}}}

 

=\frac{{1+\text{sinA}}}{{\sqrt{{\text{co}{{\text{s}}^{2}}\text{ }\!\!~\!\!\text{ A}}}}}=\frac{{1+\text{sinA}}}{{\text{cosA}}}

 

\begin{array}{*{20}{r}} {} & ~ \\ {} & {} \\ {} & {~=\frac{1}{{\text{cosA}}}+\frac{{\text{sinA}}}{{\text{cosA}}}=\text{secA}+\text{tanA}=\text{R}.\text{H}.\text{S}.} \end{array}

 

(ii)

\text{ }\!\!~\!\!\text{ L}\text{.H}\text{.S}\text{. }\!\!~\!\!\text{ =}\frac{{\sqrt{{1-\text{cos}\theta }}}}{{\sqrt{{1+\text{cos}\theta }}}}

 

=\frac{{\sqrt{{1-\text{cos}\theta }}}}{{\sqrt{{1+\text{cos}\theta }}}}\times \frac{{\sqrt{{1-\text{cos}\theta }}}}{{\sqrt{{1-\text{cos}\theta }}}}

 

[Rationalism the numerator]

 

=\frac{{1-\text{cos}\theta }}{{\sqrt{{1-\text{co}{{\text{s}}^{2}}\theta }}}}=\frac{{1-\text{cos}\theta }}{{\text{sin}\theta }}

 

\begin{array}{*{20}{r}} {} & {} \\ {} & {~=\frac{1}{{\text{sin}\theta }}-\frac{{\text{cos}\theta }}{{\text{sin}\theta }}=\text{cosec}\theta -\text{cot}\theta =\text{ }\!\!~\!\!\text{ R}.\text{H}.\text{S}.\text{ }\!\!~\!\!\text{ }} \end{array}

 

Example 4. Prove the following :


(i) \frac{{1-\text{cos}\theta }}{{\text{sin}\theta }}=\frac{{\text{sin}\theta }}{{1+\text{cos}\theta }}


(ii) \frac{{\text{sec}\theta -\text{tan}\theta }}{{\text{sec}\theta +\text{tan}\theta }}=1-2\text{sec}\theta \text{tan}\theta +2\text{ta}{{\text{n}}^{2}}\theta


Solution.

\text{(i) }\!\!~\!\!\text{ L}\text{.H}\text{.S}=\frac{{1-\text{cos}\theta }}{{\text{sin}\theta }} =\frac{{1-\text{cos}\theta }}{{\text{sin}\theta }}\times \frac{{1+\text{cos}\theta }}{{1+\text{cos}\theta }}

 

=\frac{{1-\text{co}{{\text{s}}^{2}}\theta }}{{\text{sin}\theta \left( {1+\text{cos}\theta } \right)}}=\frac{{\text{si}{{\text{n}}^{2}}\theta }}{{\text{sin}\theta \left( {1+\text{cos}\theta } \right)}}

 

=\frac{{\text{sin}\theta }}{{1+\text{cos}\theta }}=\text{ }\!\!~\!\!\text{ R}.\text{H}.\text{S}.\text{ }\!\!~\!\!\text{ }

 

(ii) L.H.S. =\frac{{\text{sec}\theta -\text{tan}\theta }}{{\text{sec}\theta +\text{tan}\theta }}

 

=\frac{{\text{sec}\theta -\text{tan}\theta }}{{\text{sec}\theta +\text{tan}\theta }}\times \frac{{\text{sec}\theta -\text{tan}\theta }}{{\text{sec}\theta -\text{tan}\theta }}


=\frac{{{{{(\text{sec}\theta -\text{tan}\theta )}}^{2}}}}{{\text{se}{{\text{c}}^{2}}\theta -\text{ta}{{\text{n}}^{2}}\theta }}

 

=\frac{{\text{se}{{\text{c}}^{2}}\theta +\text{ta}{{\text{n}}^{2}}\theta -2\text{sec}\theta \text{tan}\theta }}{1}

 

=1+\text{ta}{{\text{n}}^{2}}\theta +\text{ta}{{\text{n}}^{2}}\theta -2\text{sec}\theta \text{tan}\theta

 

=1-2\text{sec}\theta \text{tan}\theta +2\text{ta}{{\text{n}}^{2}}\theta = R.H.S.


Example 5. Prove the following:


(i) \text{se}{{\text{c}}^{4}}\theta -\text{se}{{\text{c}}^{2}}\theta =\text{ta}{{\text{n}}^{4}}\theta +\text{ta}{{\text{n}}^{2}}\theta


(ii) 2\text{se}{{\text{c}}^{2}}\theta -\text{se}{{\text{c}}^{4}}\theta -2\text{cose}{{\text{c}}^{2}}\theta +\text{cose}{{\text{c}}^{4}}\theta =\text{co}{{\text{t}}^{4}}\theta -\text{ta}{{\text{n}}^{4}}\theta


(iii) \text{si}{{\text{n}}^{\text{*}}}\theta -\text{co}{{\text{s}}^{8}}\theta =\left( {\text{si}{{\text{n}}^{2}}\theta -\text{co}{{\text{s}}^{2}}\theta } \right)\left( {1-2\text{si}{{\text{n}}^{2}}\theta \text{co}{{\text{s}}^{2}}\theta } \right)


(iv) \text{se}{{\text{c}}^{6}}0-\text{ta}{{\text{n}}^{6}}\theta =3\text{ta}{{\text{n}}^{2}}\theta \text{se}{{\text{c}}^{2}}\theta +1


(v) 2\left( {\text{si}{{\text{n}}^{6}}\theta +\text{co}{{\text{s}}^{6}}\theta } \right)-3\left( {\text{si}{{\text{n}}^{4}}\theta +\text{co}{{\text{s}}^{4}}\theta } \right)+1=0.


Solution. (i) L.H.S. =\text{se}{{\text{c}}^{4}}\theta -\text{se}{{\text{c}}^{2}}\theta

 

=\text{se}{{\text{c}}^{2}}\theta \left( {\text{se}{{\text{c}}^{2}}\theta -1} \right)

 

\begin{array}{*{20}{r}} {} & {} \\ {} & {~=\left( {1+\text{ta}{{\text{n}}^{2}}\theta } \right)\text{ta}{{\text{n}}^{2}}\theta =\text{ta}{{\text{n}}^{2}}\theta +\text{ta}{{\text{n}}^{4}}\theta =\text{ }\!\!~\!\!\text{ R}.\text{H}.\text{S}.\text{ }\!\!~\!\!\text{ }} \end{array}

 

\text{ }\!\!~\!\!\text{ (ii) }\!\!~\!\!\text{ L}\text{.H}\text{.S}\text{. }\!\!~\!\!\text{ =}2\text{se}{{\text{c}}^{2}}\theta -\text{se}{{\text{c}}^{4}}\theta -2\text{cose}{{\text{c}}^{2}}\theta +\text{cose}{{\text{c}}^{4}}\theta

 

=2\left( {1+\text{ta}{{\text{n}}^{2}}\theta } \right)-{{\left( {1+\text{ta}{{\text{n}}^{2}}\theta } \right)}^{2}}-2\left( {1+\text{co}{{\text{t}}^{2}}\theta } \right)+{{\left( {1+\text{co}{{\text{t}}^{2}}\theta } \right)}^{2}}

 

=\text{ }\!\!~\!\!\text{ }2+2\text{ta}{{\text{n}}^{2}}\theta -\left( {1+\text{ta}{{\text{n}}^{4}}\theta +2\text{ta}{{\text{n}}^{2}}\theta } \right)-2-2\text{co}{{\text{t}}^{2}}\theta +1

 

=\text{ }\!\!~\!\!\text{ }2+2\text{ta}{{\text{n}}^{2}}\theta -1-\text{ta}{{\text{n}}^{4}}\theta +2\text{co}{{\text{t}}^{2}}\theta

 

=\text{ }\!\!~\!\!\text{ co}{{\text{t}}^{4}}\theta -\text{ta}{{\text{n}}^{4}}\theta =\text{ }\!\!~\!\!\text{ R}.\text{H}.\text{S}

 

\text{(iii) }\!\!~\!\!\text{ L}\text{.H}\text{.S}\text{. }\!\!~\!\!\text{ = }\!\!~\!\!\text{ si}{{\text{n}}^{8}}\theta -\text{co}{{\text{s}}^{8}}\theta =\left( {\text{si}{{\text{n}}^{4}}\theta -\text{co}{{\text{s}}^{4}}\theta } \right)\left( {\text{si}{{\text{n}}^{4}}\theta +\text{co}{{\text{s}}^{4}}\theta } \right)

 

=~\left[ {\left( {\text{si}{{\text{n}}^{2}}\theta +\text{co}{{\text{s}}^{2}}\theta } \right)\left( {\text{si}{{\text{n}}^{2}}\theta -\text{co}{{\text{s}}^{2}}\theta } \right)} \right]\left[ {{{{\left( {\text{si}{{\text{n}}^{2}}\theta +\text{co}{{\text{s}}^{2}}\theta } \right)}}^{2}}} \right.

 

=~\left( {\text{si}{{\text{n}}^{2}}\theta -\text{co}{{\text{s}}^{2}}\theta } \right)\left( {1-2\text{si}{{\text{n}}^{2}}\theta \text{co}{{\text{s}}^{2}}\theta } \right)=\text{ }\!\!~\!\!\text{ R}.\text{H}.\text{S}.

 

\text{(iv) }\!\!~\!\!\text{ L}\text{.H}\text{.S}\text{. }\!\!~\!\!\text{ = }\!\!~\!\!\text{ se}{{\text{c}}^{6}}\theta -\text{ta}{{\text{n}}^{6}}\theta ={{\left( {\text{se}{{\text{c}}^{2}}\theta } \right)}^{3}}-{{\left( {\text{ta}{{\text{n}}^{2}}\theta } \right)}^{3}}

 

=~\left( {\text{se}{{\text{c}}^{2}}\theta -\text{ta}{{\text{n}}^{2}}\theta } \right)\left( {\text{se}{{\text{c}}^{4}}\theta +\text{se}{{\text{c}}^{2}}\theta \text{ta}{{\text{n}}^{2}}\theta +\text{ta}{{\text{n}}^{4}}\theta } \right)

 

=~\left( {\text{se}{{\text{c}}^{2}}\theta -\text{ta}{{\text{n}}^{2}}\theta } \right)\left[ {{{{\left( {\text{se}{{\text{c}}^{2}}\theta -\text{ta}{{\text{n}}^{2}}\theta } \right)}}^{2}}+\left( {3\text{se}{{\text{c}}^{2}}\theta \text{ta}{{\text{n}}^{2}}\theta } \right)} \right]

 

=\text{ }\!\!~\!\!\text{ }1+3\text{se}{{\text{c}}^{2}}\theta \text{ta}{{\text{n}}^{2}}\theta =\text{ }\!\!~\!\!\text{ R}.\text{H}.\text{S}.\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\left[ {\because \text{se}{{\text{c}}^{2}}\theta -\text{ta}{{\text{n}}^{2}}\theta =1} \right]

 

=2\left( {\text{co}{{\text{s}}^{6}}\theta +\text{si}{{\text{n}}^{6}}\theta } \right)-3\left( {\text{co}{{\text{s}}^{4}}\theta +\text{si}{{\text{n}}^{4}}\theta } \right)+1

 

=\text{ }\!\!~\!\!\text{ }2\left[ {{{{\left( {\text{co}{{\text{s}}^{2}}\theta +\text{si}{{\text{n}}^{2}}\theta } \right)}}^{3}}-3\text{co}{{\text{s}}^{2}}\theta \text{si}{{\text{n}}^{2}}\theta \left( {\text{co}{{\text{s}}^{2}}\theta +\text{si}{{\text{n}}^{2}}\theta } \right)} \right]

 

\left. {=\text{ }\!\!~\!\!\text{ }-3\left[ {{{{\left( {\text{co}{{\text{s}}^{2}}\theta +\text{si}{{\text{n}}^{2}}\theta } \right)}}^{2}}-2\text{co}{{\text{s}}^{2}}\theta \text{si}{{\text{n}}^{2}}\theta } \right)} \right]+1

 

=\text{ }\!\!~\!\!\text{ }2\left[ {1-3\text{co}{{\text{s}}^{2}}\theta \text{si}{{\text{n}}^{2}}\theta } \right]-3\left[ {1-2\text{co}{{\text{s}}^{2}}\theta \text{si}{{\text{n}}^{2}}\theta } \right]+1

 

\left[ {\because \text{si}{{\text{n}}^{2}}\theta +\text{co}{{\text{s}}^{2}}\theta =1} \right]

 

=\text{ }\!\!~\!\!\text{ }2-6\text{co}{{\text{s}}^{2}}\theta \text{si}{{\text{n}}^{2}}\theta -3+6\text{si}{{\text{n}}^{2}}\theta \text{co}{{\text{s}}^{2}}\theta +1=0=\text{ }\!\!~\!\!\text{ R}.\text{H}.\text{S}.

 

Example 6. Prove the following :


(i) \frac{{\text{cos}x}}{{1-\text{sin}x}}+\frac{{1-\text{sin}x}}{{\text{cos}x}}=2\text{sec}x


(ii) \frac{{\text{cot}A+\text{tan}B}}{{\text{cot}B+\text{tan}A}}=\text{cot}A\text{tan}B


(iii) \frac{{\text{cos}\theta }}{{1-\text{tan}\theta }}+\frac{{\text{sin}\theta }}{{1-\text{cot}\theta }}=\text{cos}\theta +\text{sin}\theta .


(iv) \frac{1}{{\text{sec}\theta -\text{tan}\theta }}-\frac{1}{{\text{cos}\theta }}=\frac{1}{{\text{cos}\theta }}-\frac{1}{{\text{sec}\theta +\text{tan}\theta }}


Solution. (i) L.H.S. =\frac{{\text{cos}x}}{{1-\text{sin}x}}+\frac{{1-\text{sin}x}}{{\text{cos}x}}

 

=\frac{{\text{co}{{\text{s}}^{2}}x+{{{(1-\text{sin}x)}}^{2}}}}{{\text{cos}x\left( {1-\text{sin}x} \right)}}=\frac{{\text{co}{{\text{s}}^{2}}x+1+\text{si}{{\text{n}}^{2}}x-2\text{sin}x}}{{\text{cos}x\left( {1-\text{sin}x} \right)}}

 

=\frac{{\text{co}{{\text{s}}^{2}}x+{{{(1-\text{sin}x)}}^{2}}}}{{\text{cos}x\left( {1-\text{sin}x} \right)}}=\frac{{\text{co}{{\text{s}}^{2}}x+1+\text{si}{{\text{n}}^{2}}x-2\text{sin}x}}{{\text{cos}x\left( {1-\text{sin}x} \right)}}

 

=\frac{{1+\left( {\text{si}{{\text{n}}^{2}}x+\text{co}{{\text{s}}^{2}}x} \right)-2\text{sin}x}}{{\text{cos}x\left( {1-\text{sin}x} \right)}}=\frac{{2-2\text{sin}x}}{{\text{cos}x\left( {1-\text{sin}x} \right)}}

 

=\frac{{2\left( {1-\text{sin}x} \right)}}{{\text{cos}x\left( {1-\text{sin}x} \right)}}=\frac{2}{{\text{cos}x}}=2\text{sec}x=\text{ }\!\!~\!\!\text{ R}.\text{H}.\text{S}.

 

\text{(ii) }\!\!~\!\!\text{ L}\text{.H}\text{.S}\text{.}~=\frac{{\text{cotA}+\text{tanB}}}{{\text{cotB}+\text{tanA}}}\text{ }\!\!~\!\!\text{ }

 

=\frac{{\frac{1}{{\frac{{\text{tanA}}}{1}+\text{tanB}}}\text{tanB}+\frac{{1+\text{tanAtanB}}}{{\frac{{\text{tanA}}}{{1+\text{tanAtanB}}}}}}}{{\text{tanB}}}

 

=\frac{{\text{tanB}}}{{\text{tanA}}}=\frac{1}{{\text{tanA}}}\cdot \text{tanB}=\text{cot}\mathbf{A}\text{tan}\mathbf{B}=\text{ }\!\!~\!\!\text{ R}.\text{H}.\text{S}.\text{ }\!\!~\!\!\text{ }

 

\text{(iii) }\!\!~\!\!\text{ L}\text{.H}\text{.S}~=\frac{{\text{cos}\theta }}{{1-\text{tan}\theta }}+\frac{{\text{sin}\theta }}{{1-\text{cot}\theta }}\text{.}

 

=\frac{{\text{cos}\theta }}{{1-\frac{{\text{sin}\theta }}{{\text{cos}\theta }}}}+\frac{{\text{sin}\theta }}{{1-\frac{{\text{cos}\theta }}{{\text{sin}\theta }}}}

 

=\frac{{\text{co}{{\text{s}}^{2}}\theta }}{{\text{cos}\theta -\text{sin}\theta }}+\frac{{\text{si}{{\text{n}}^{2}}\theta }}{{\text{sin}\theta -\text{cos}\theta }}

 

=\frac{{\text{co}{{\text{s}}^{2}}\theta }}{{\text{cos}\theta -\text{sin}\theta }}-\frac{{\text{si}{{\text{n}}^{2}}\theta }}{{\text{cos}\theta -\text{sin}\theta }}=\frac{{\text{co}{{\text{s}}^{2}}\theta -\text{si}{{\text{n}}^{2}}\theta }}{{\text{cos}\theta -\text{sin}\theta }}

 

=\frac{{\left( {\text{cos}\theta -\text{sin}\theta } \right)\left( {\text{cos}\theta +\text{sin}\theta } \right)}}{{\text{cos}\theta -\text{sin}\theta }}=\text{cos}\theta +\text{sin}\theta = R.H.S.

 


(iv) \frac{1}{{\text{sec}\theta -\text{tan}\theta }}-\frac{1}{{\text{cos}\theta }}=\frac{1}{{\text{cos}\theta }}-\frac{1}{{\text{sec}\theta +\text{tan}\theta }}


\frac{1}{{\text{sec}\theta -\text{tan}\theta }}+\frac{1}{{\text{sec}\theta +\text{tan}\theta }}=\frac{2}{{\text{cos}\theta }}

 

or \text{ }\!\!~\!\!\text{ } L.H.S. =\frac{1}{{\text{sec}\theta -\text{tan}\theta }}+\frac{1}{{\text{sec}\theta +\text{tan}\theta }}


=\frac{{\text{sec}\theta +\text{tan}\theta +\text{sec}\theta -\text{tan}\theta }}{{\left( {\text{sec}\theta -\text{tan}\theta } \right)\left( {\text{sec}\theta +\text{tan}\theta } \right)}}

 

=\frac{{2\text{sec}\theta }}{{\text{se}{{\text{c}}^{2}}\theta -\text{ta}{{\text{n}}^{2}}\theta }}=\frac{{2\text{sec}\theta }}{1}=2\text{sec}\theta =\frac{2}{{\text{cos}\theta }}= R.H.S.

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