Trigonometry Class 10

    \[ % Generated by GrindEQ Word-to-LaTeX \documentclass{article} %%% use \documentstyle for old LaTeX compilers \usepackage[english]{babel} %%% 'french', 'german', 'spanish', 'danish', etc. \usepackage{amssymb} \usepackage{amsmath} \usepackage{txfonts} \usepackage{mathdots} \usepackage[classicReIm]{kpfonts} \usepackage[dvips]{graphicx} %%% use 'pdftex' instead of 'dvips' for PDF output % You can include more LaTeX packages here \begin{document} %\selectlanguage{english} %%% remove comment delimiter ('%') and select language if required \begin{enumerate} \item  $\frac{1}{secA+tanA}-\frac{1}{cosA}=\frac{1}{cosA}-\frac{1}{secA-tanA}$ \end{enumerate} \noindent LHS \[\frac{1}{secA+tanA}-\frac{1}{cosA}\]

    \[=\frac{1}{secA+tanA}\times \frac{secA-tanA}{secA-tanA}-\frac{1}{cosA}\]

(Multiplying and dividing by secA-tanA, we get)

    \[\frac{secA-tanA\ }{{sec}^2A-{tan}^{2\ }A}-secA   \left(using\ \frac{1}{cosA}=secA\right)\]

    \[=\frac{secA-tanA\ }{1}-secA \left\{1+{tan}^2\theta ={sec}^2\theta \ therefore\ {sec}^2\theta -{tan}^2\theta =1\right\}\]

    \[=secA\ --tanA-secA\]

    \[=\frac{1}{cosA}-(tanA+secA)\]

    \[=\frac{1}{cosA}-\left(secA+tanA\right)\times \frac{secA-tanA}{ \begin{array}{c} secA-ta\textrm{?}\textrm{?}A \\  \end{array} }\left\{\mathrm{Multiplying\ and\ dividing\ by\ secA\ }--\mathrm{tanA}\right\}\]

    \[=\frac{1}{cosA}-\frac{{sec}^2A-{tan}^2A}{{sec}^2A}=\frac{1}{cosA}-\frac{1}{secA-tanA}=RHS\]

 

\begin{enumerate}

\item  {tan}^2A+{cot}^2A={sec}^2A+{cosec}^2A-2

\end{enumerate}

 

\noindent LHS 

    \[{tan}^2A+{cot}^2A={sec}^2A+{cosec}^2A-1\]

    \[{=sec}^2A-1+{cosec}^2A-1\]

    \[={sec}^2A+{cosec}^2A-2\]

    \[=\frac{1}{{cos}^{2\ }A}+\frac{1}{{sin}^{2\ }A}-\ 2\]

    \[=\frac{{sin}^{2\ }A+{cos}^2A}{{cos}^{2\ }A{sin}^2A}-2\]

    \[=\frac{1}{{cos}^2A{sin}^2A}-2\ \]

    \[=\frac{1}{{cos}^2A}\ \times \ \frac{1}{{sin}^2A}-2\]

    \[={sec}^2A{cosec}^2A=R.H.S.\]

 

 

\begin{enumerate}

\item  \frac{tanA}{1+secA}-\frac{tanA}{1-secA}=2cosecA

\end{enumerate}

 

\noindent L.H.S.

    \[\frac{tanA}{1+secA}-\frac{tanA}{1-secA}\]

    \[=\frac{\frac{sinA}{cosA}}{1+\frac{1}{cosA}}-\frac{\frac{sinA}{cosA}}{1-\frac{1}{cosA}}\]

    \[=\frac{\frac{sinA}{cosA}}{\frac{cosA+1}{cosA}}-\frac{\frac{sinA}{cosA}}{\frac{cosA-1}{cosA}}\]

    \[=\ \frac{sinA}{cosA}\times \frac{cosA}{cosA+1}-\ \frac{sinA}{cosA}\times \frac{cosA}{cosA-1}\]

    \[=\frac{sinAcosA-sinA-sinAcosA-sinA\ }{{cos}^2A-1}\]

    \[=\ \frac{2sinA}{\left(1-{cos}^2A\right)}\]

    \[=\ \frac{2sinA}{{sin}^2A}\]

    \[=\frac{2}{sinA}\]

    \[=2cosecA=R.H.S\]

    \[Alternate\ Method\ :\]

    \[L.H.S.\]

    \[\frac{tanA}{1+secA}-\ \frac{tanA}{1-secA}\]

    \[=\ \frac{tanA-tanAsecA-t\textrm{?}\textrm{?}nA-tanAsecA}{\left(1+secA\right)\ \left(1-secA\right)}\]

    \[=\ \frac{-2tanAsecA}{1-{sec}^2A}\theta =cosec\theta =R.H.S.\]

 

\begin{enumerate}

\item  Prove that :\frac{cos\theta }{cosec\theta +1\ }+\frac{cos\theta }{cosec\theta -1}=2tan\theta

\end{enumerate}

 

\noindent L.H.S.

    \[\frac{cos\theta }{cosec\theta +1\ }+\frac{cos\theta }{cosec\theta -1}\]

    \[\mathrm{=cos}\theta \left(\frac{1}{cosec\theta +1}+\frac{1}{cosec\theta -1}\right)\]

    \[=cos\theta \left(\frac{cosec\theta -1+cosec\theta +1}{\left(cosec\theta +1\right)\left(cosec\theta -1\right)}\right)\]

    \[=cos\theta \ \times \frac{2cosec\theta }{{cosec}^2\theta -1}\]

    \[=\frac{cos\theta \left(2cosec\theta \right)}{{cot}^2\theta }\]

    \[=\frac{cos\theta \times 2\times \frac{1}{sin\theta }}{\frac{{cos}^2\theta }{{sin}^2\theta }}\]

    \[=\ \frac{2cos\theta }{sin\theta }\times \ \frac{{sin}^2\theta }{{cos}^2\theta }\]

    \[=2tan\theta =R.H.S.\]

 

\begin{enumerate}

\item  Prove that   :     1+{tan}^2A+\left(1+\frac{1}{{tan}^2A}\right)=\frac{1}{{sin}^2A-{sin}^4A}

\end{enumerate}

 

\noindent L.H.S

    \[\mathrm{\ 1+}{tan}^2A+\left(1+\frac{1}{{tan}^2A}\right)\]

    \[={sec}^2A+\left(1+{cot}^2A\right)\]

    \[={sec}^2A+{cosec}^2A\]

    \[=\frac{1}{{cos}^2A}+\frac{1}{{sin}^2A}\]

    \[=\frac{{sin}^2A+{cos}^2A}{{cos}^2A\ {sin}^2A}\]

    \[=\frac{1}{\left(1-{sin}^2A\right){sin}^2A}\]

    \[=\frac{1}{{sin}^2A-\ {sin}^4A}=R.H.S\]

 

\begin{enumerate}

\item   PROVE THAT      :        {sin}^2A{cos}^2B-\ {cos}^2A{sin}^2B=\ {sin}^2A{sin}^2B

\end{enumerate}

 

\noindent L.H.S.

    \[{sin}^2A{cos}^2B-\ {cos}^2A{sin}^2B\]

    \[=\ \ {sin}^2A\left(1-{sin}^2B\right)-\left(1-{sin}^2A\right){sin}^2B\]

    \[={sin}^2A-{sin}^2A{sin}^2B-{sin}^2B+{sin}^2A{sin}^2B\ \]

    \[={sin}^2A-{sin}^2B=R.H.S.\]

 

\begin{enumerate}

\item 

 

\item  Prove that :  \frac{cotA+tanB}{cotB+tanA}=cotAtanB

\end{enumerate}

 

\noindent L.H.S.

    \[\frac{cotA+tanB}{cotB+tanA}\]

    \[=\ \frac{\frac{cosA}{sinA}+\frac{sinB}{cosB}}{\frac{cosB}{sinB}+\frac{sinA}{cosA}}\]

    \[=\ \frac{\frac{cosAcosB+sinAsinB}{sinAcosB}}{\frac{c\textrm{?}\textrm{?}sBcosA+sinAsinB}{sinBcosA}}\]

    \[=\ \frac{\left(cosAcosB+sinAsinB\right)}{sinAcosB}\ \times \frac{sinBcosA}{cosAcosB+sinAsinB}\]

 

    \[=\frac{sinBcosA}{cosBsinA}=tanBcotA\]

    \[=cotAtanB=R.H.S.\]

 

 

\begin{enumerate}

\item  Prove that: \frac{tanA+tanB}{cotA+cotB}=tanAtanB

\end{enumerate}

 

\noindent L.H.S.

    \[\frac{tanA+tanB}{cotA+cotB}\]

    \[=\ \frac{\frac{sinA}{cosA}+\frac{sinB}{cosB}}{\frac{cosA}{sinB}+\frac{cosB}{sinB}}\]

    \[=\frac{\frac{sinAcosB+cosAsinB}{cosAcosB}}{\frac{cosAsinB+sinAcosB}{sinAsinB}}\]

    \[=\frac{sinAsinB}{cosAcosB}\]

    \[=tanAtanB=R.H.S.\]

 

 

\begin{enumerate}

\item  Prove that: {cot}^2A{cosec}^2B-{cot}^2B{cosec}^2A={cot}^2A-{cot}^{ \begin{array}{c}  2 \\   \end{array}  }B

\end{enumerate}

 

\noindent L.H.S.

    \[{cot}^2A{cosec}^2B-{cot}^2B{cosec}^2A\]

    \[=\frac{{cos}^2A}{{sin}^2A}\ \times \frac{1}{{sin}^2B}-\frac{{cos}^2B}{{sin}^2B}\ \times \frac{1}{{sin}^2A}\]

    \[=\ \frac{{cos}^2A-{cos}^2B}{{sin}^2A{sin}^2B}\]

    \[=\frac{{cos}^2A-{cos}^2B+{cos}^2A{cos}^2B-{cos}^2A{cos}^2B}{{sin}^2A{sin}^2B}\ \]

 

 

\noindent Adding and subtracting {cos}^2A+{cos}^2B in numerator, we get

 

\noindent

    \[\frac{{cos}^2A-{cos}^2A{cos}^2B-{cos}^2B+{cos}^2A{cos}^2B}{{sin}^2A{sin}^2B}\]

    \[=\frac{{cos}^2A\left(1-{cos}^2B\right)-{cos}^2B\left(1-{cos}^2A\right)\ }{{sin}^2A{sin}^2B}\]

    \[=\frac{{cos}^2A{sin}^2B}{{sin}^2A{sin}^2B}-\ \frac{{cos}^2B{sin}^2A}{{sin}^2A{sin}^2B}\]

={cot}^2A-{cot}^2B = R.H.S.

 

\begin{enumerate}

\item  Prove that   :  {tan}^2A{sec}^2B-\ {sec}^2A{tan}^2B=\ {tan}^2A-\ {tan}^2B

\end{enumerate}

 

\noindent L.H.S.\ \ \ \ {tan}^2A\left(1-{tan}^2B\right)-\ \left(1+{tan}^2A\right){tan}^2B

    \[=\ {tan}^2A+{tan}^2A{tan}^2B- {tan}^2B-{tan}^2A{tan}^2B\]

={tan}^2A-\ {tan}^2B=\R.H.S.

 

\begin{enumerate}

\item  If x=asec\theta +bsec\theta \ and\ y=atan\theta +bsec\theta

\end{enumerate}

 

\noindent Prove that  :  x^2-y^2=\ a^2-\ b^2

 

\noindent L.H.S.

    \[x^2-y^2={\left(asec\theta +bsec\theta \right)}^2-\ {\left(atan\theta +bsec\theta \right)}^2\]

    \[=a^2{sec}^2\theta +b^2{tan}^2\theta +2ab\ sec\theta \ tan\theta -\ a^2{tan}^2\theta +b^2{sec}^2\theta +2ab\ tan\theta \ sec\theta \]

    \[=a^2{sec}^2\theta -a^2{tan}^2\theta +b^2{tan}^2\theta -b^2{sec}^2\theta \]

    \[=\ a^2\left({sec}^2\theta -{tan}^2\theta \right)-b^2\left({sec}^2\theta -{tan}^2\theta \right)\]

    \[=a^2\left(1\right)-b^2\left(1\right)=\ a^2-b^{2\ }=R.H.S.\]

 

\begin{enumerate}

\item  3sin\theta +5cos\theta =5\dots \left(1\right)

\end{enumerate}

 

\noindent Let   5sin\theta -3cos\theta =x\dots \left(2\right)

 

\noindent Squaring and adding \eqref{GrindEQ__1_} and \eqref{GrindEQ__2_}, we get

    \[{\Rightarrow \left(3sin\theta +5cos\theta \right)}^2+{\left(5sin\theta -3cos\theta \right)}^2=5^2+x^2\]

    \[\Rightarrow 9{sin}^2\theta +25{cos}^2\theta +30\ sin\theta \ cos\theta +25{sin}^2\theta +9{cos}^2\theta -30\ sin\theta \ cos\theta =25+x^2\]

    \[\Rightarrow 9{sin}^2\theta +25{sin}^2\theta +25{cos}^2\theta +9{cos}^2\theta -25=x^2\]

    \[\Rightarrow 34{sin}^2\theta +34{cos}^2\theta -25=x^2\]

    \[\Rightarrow 34\left({sin}^2\theta +{cos}^2\theta \right)-25=x^{2\ }\]

    \[{\Rightarrow x}^2=34-25=9\]

    \[\Rightarrow x=\sqrt{9}=3\]

Therefore 5sin\theta -3cos\theta =3

 

\begin{enumerate}

\item  If  cosec\theta +cot\theta =m;\ \ \ cosec\theta -cot\theta =n\ then\ prove\ that\ mn=1

\end{enumerate}

 

\noindent R.H.S.

    \[mn\]

    \[=\left(cosec\theta +cot\theta \right)\left(cosec\theta -cot\theta \right)\]

    \[={cosec}^2\theta -{cot}^2\theta  =1\]

    \[\frac{{tan}^3\theta }{1+{tan}^2\theta }+\frac{{cot}^3\theta }{1+{cot}^2\theta }=\frac{{tan}^3\theta }{1+{tan}^2\theta }+\frac{{cot}^3\theta }{1+{cot}^2\theta }=\]

 

\begin{enumerate}

\item 

    \[\frac{{tan}^3\theta }{1+{tan}^2\theta }+\frac{{cot}^3\theta }{1+{cot}^2\theta }=sec\theta cosec\theta -2sin\theta cos\theta \]

\end{enumerate}

L.H.S.

    \[\frac{{tan}^3\theta }{1+{tan}^2\theta }+\frac{{cot}^3\theta }{1+{cot}^2\theta }\]

    \[=\frac{{tan}^3\theta }{{sec}^2\theta }+\frac{{cot}^3\theta }{{cosec}^2\theta }\]

    \[=\frac{\frac{{sin}^3\theta }{{cos}^3\theta }}{\frac{1}{{cos}^2\theta }}+\frac{\frac{{cos}^3\theta }{{sin}^3\theta }}{\frac{1}{{sin}^2\theta }}\]

    \[=\ \frac{{sin}^3\theta }{{cos}^3\theta }\times {cos}^2\theta +\frac{{cos}^3\theta }{{sin}^3\theta }\times {sin}^2\theta \]

    \[=\frac{{sin}^3\theta }{cos\theta }+\frac{{cos}^3\theta }{sin\theta }\]

    \[=\frac{{sin}^4\theta +{cos}^4\theta }{cos\theta sin\theta }\]

    \[=\ \frac{{\left({sin}^2\theta \right)}^2+{\left({cos}^2\theta \right)}^2}{cos\theta sin\theta }\]

    \[=\frac{{\left({sin}^2\theta \right)}^2+{\left({cos}^2\theta \right)}^2+2{sin}^2\theta {cos}^2\theta -2{sin}^2\theta {cos}^2\theta }{cos\theta \ sin\theta }  \left\{adding\ and\ subtracting\ 2{sin}^2\theta {cos}^2\theta \right\}\]

    \[=\frac{{\left({sin}^2\theta +{cos}^2\theta \right)}^2-2{sin}^2\theta {cos}^2\theta }{sin\theta cos\theta }\ \left\{{Using\ a}^2+b^2+2ab=\ {\left(a+b\right)}^2\right\}\]

    \[= \frac{1}{sin\theta cos\theta }-\frac{2{sin}^2\theta {cos}^2\theta }{sin\theta cos\theta }\]

    \[=\ \frac{1}{sin\theta }\frac{1}{cos\theta }-\ 2sin\theta cos\theta \]

    \[=cosec\theta sec\theta -2sin\theta cos\theta \]

    \[=sec\theta cosec\theta -2sin\theta cos\theta =R.H.S.\]

    \[62.\]

    \[T_n=\ {sin}^n\theta +{cos}^n\theta ;\ prove\ \frac{T_{3\ -{\ T}_5}}{T_1}=\ \frac{T_5-\ T_7}{T_3}\]

Whereas T_1=sin\theta cos\theta ;\ T_3={sin}^3\theta +{cos}^3\theta ;T_5={sin}^5\theta +{cos}^5\theta ;T_7=\ {sin}^7\theta {cos}^7\theta

 

\noindent L.H.S:

    \[\frac{T_{3\ }-T_5}{T_1}\]

    \[=\ \frac{{sin}^3\theta +{cos}^3\theta -\left({sin}^5\theta +{cos}^5\theta \right)}{sin\theta +cos\theta }\]

    \[=\frac{{sin}^3\theta +{cos}^3\theta -{sin}^5\theta -{cos}^5\theta }{sin\theta +cos\theta }\]

    \[=\frac{{sin}^3\theta -{sin}^5\theta +{cos}^3\theta -{cos}^5\theta }{sin\theta +cos\theta }\]

    \[=\frac{{sin}^3\theta \left(1-{sin}^2\theta \right)+{cos}^3\theta \left(1-{cos}^2\theta \right)}{sin\theta +cos\theta }\]

    \[=\frac{{sin}^3\theta \left(1-{sin}^2\theta \right)+{cos}^3\left(1-{cos}^2\theta \right)}{sin\theta +cos\theta }\]

    \[=\frac{{sin}^3\theta {cos}^2\theta +{cos}^3\theta {sin}^2\theta }{sin\theta cos\theta }\]

    \[=\frac{{sin}^2\theta {cos}^2\theta \left(sin\theta cos\right)}{sin\theta cos\theta }\]

    \[={sin}^2\theta {cos}^2\theta \dots \left(1\right)\]

R.H.S.

    \[\frac{T_5-T_7}{T_3}\]

    \[=\frac{{sin}^5\theta +{cos}^5\theta -\ \left({sin}^7\theta +{cos}^7\theta \right)}{{sin}^3\theta +{cos}^3\theta }\]

    \[=\frac{{sin}^5\theta +{cos}^5\theta -{sin}^7\theta -{cos}^7\theta }{{sin}^3\theta +{cos}^3\theta }\]

    \[=\frac{{sin}^5\theta -{sin}^7\theta +{cos}^5\theta -{cos}^7\theta }{{sin}^3\theta +{cos}^3\theta }\]

    \[=\frac{{sin}^5\theta \left(1-{sin}^2\theta \right)+{cos}^5\theta \left(1-{cos}^2\theta \right)}{{sin}^3\theta +{cos}^{3\theta }}\]

    \[=\frac{{sin}^5\theta {cos}^2\theta +{cos}^5\theta {sin}^2\theta }{{sin}^3\theta +{cos}^3\theta }\]

    \[=\frac{{sin}^2\theta {cos}^2\theta \left({sin}^3\theta {+cos}^3\theta \right)}{{sin}^3\theta +{cos}^3\theta }\]

    \[={sin}^2\theta {cos}^2\theta \dots \left(2\right)\]

From \eqref{GrindEQ__1_} and \eqref{GrindEQ__2_}, L.H.S = R.H.S.

    \[63.\]

L.H.S.

    \[{\left(tan\theta +\frac{1}{cos\theta }\right)}^2+{\left(tan\theta -\frac{1}{cos\theta }\right)}^2\]

    \[={tan}^2\theta +\frac{1}{{cos}^2\theta }+2tan\theta \frac{1}{cos\theta }+{tan}^2\theta +\frac{1}{{cos}^2\theta }-2tan\theta \frac{1}{cos\theta }\]

    \[=2{tan}^2\theta +\frac{2}{{cos}^2\theta }\]

    \[=2\frac{{sin}^2\theta }{{cos}^2\theta }+\frac{2}{{cos}^2\theta }\]

    \[=2\left(\frac{{sin}^2\theta +1}{{cos}^2\theta }\right)\]

    \[=2\left(\frac{1+{sin}^2\theta }{1-{sin}^2\theta }\right)=R.H.S.\]

 

 

 

\end{document}

 

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