Question

If cos alpha /cos beta = m & cos alpha / sin beta =n show that (m^2+ n^2)cos beta =n^2

Answer 1

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Answer 2

$\huge{\color{t}{\textsf{\textbf {\underline{\underline{ Question : }}}}}}$

If cos alpha /cos beta = m & cos alpha / sin beta =n show that (m^2+ n^2)cos beta =n^2

$\huge\mathbb\fcolorbox{Green}{violet}{♡ᎪղՏωᎬя᭄}$

Gɪᴠᴇɴ :

$✯ \begin{gathered}\bf{\sin{\alpha}\:+\:\sin{\beta}\:=\:m} \\ \end{gathered}$

Aɴᴅ,

$✯ \begin{gathered}\bf{\cos{\alpha}\:+\:\cos{\beta}\:=\:n} \\ \end{gathered}$

Tᴏ Pʀᴏᴏғ :

$\begin{gathered}\bf{\cos(\alpha\:-\:\beta)\:=\:\dfrac{1}{2}\:(m^2\:+\:n^2\:-\:2)} \\ \end{gathered}$

Sᴏᴍᴇ Pʀᴏᴘᴇʀᴛɪᴇs :

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{\cos{(\alpha\:-\:\beta)}\:=\:\cos\alpha\:\cos\beta\:+\:\sin\alpha\:\sin\beta}}}}}} \\ \end{gathered}$

$\begin{gathered}\pink\bigstar\:\:{\underline{\blue{\boxed{\bf{\purple{\sin^2\theta\:+\:\cos^2\theta\:=\:1\:}}}}}} \\ \end{gathered}$

$➙ \begin{gathered}\bf{\dfrac{1}{2}\:(m^2\:+\:n^2\:-\:2)} \\ \end{gathered}$

☆ Now we calculate the value of 'm² + n²'.

$\begin{gathered}:\rightarrow\:\bf{(\sin{\alpha}\:+\:\sin{\beta})^2\:+\:(\cos{\alpha}\:+\:\cos{\beta})^2} \\ \end{gathered}$

$\begin{gathered}:\rightarrow\:\bf{\Big(\sin^2{\alpha}\:+\:2\:{\sin\alpha}\:{\sin\beta}\:+\:\sin^2{\beta}\Big)\:+\:\Big(\cos^2{\alpha}\:+\:2\:{\cos\alpha}\:{\cos\beta}\:+\:\cos^2{\beta}\Big)} \\ \end{gathered}$

$\begin{gathered}:\rightarrow\:\bf{(\sin^2{\alpha}\:+\:\cos^2{\alpha})\:+\:(\sin^2{\beta}\:+\:\cos^2{\beta})\:+\:(2\:{\sin\alpha}\:{\sin\beta}\:+\:2\:{\cos\alpha}\:{\cos\beta})\:} \\ \end{gathered}$

$\begin{gathered}:\rightarrow\:\bf{1\:+\:1\:+\:2\:(\sin\alpha\:\sin\beta\:+\:\cos\alpha\:\cos\beta)\:} \\ \end{gathered} :→1+1+2(sinαsinβ+cosαcosβ)$

$\begin{gathered}:\rightarrow\bf{2\:+\:2\:\cos(\alpha\:-\:\beta)\:} \\ \end{gathered}$

☆ Now putting the value of 'm² + n²' in R.H.S, we get

$➙ \begin{gathered}\bf{\dfrac{1}{2}\:\Big\{2\:+\:2\:\cos(\alpha\:-\:\beta)\:-\:2\Big\}} \\ \end{gathered}$

$➙ \begin{gathered}\bf{\dfrac{1}{2}\times{2\:\cos(\alpha\:-\:\beta)}} \\ \end{gathered}$

$➙ \begin{gathered}\bf{\underline{\green{\boxed{\pink{\cos(\alpha\:-\:\beta)}}}}}\:(L.H.S) \\ \end{gathered}$

$\Large\bf{Therefore,}$

★ L.H.S = R.H.S [Hence Proved]