Question

If tanA=ntanB and SinA = msinB, prove that (see drawing)

Answer 1

Appropriate Question:

If tanA = ntanB and sinA = msinB, prove that ${cos}^{2}A = \frac{ {m}^{2} - 1 }{ {n}^{2} - 1}$

$\large\underline{\sf{Solution-}}$

Given that,

$\sf \: tanA = n \: tanB \: \implies\sf \: \boxed{\sf \: n = \dfrac{tanA}{tanB} \: }$

Also, given that

$\sf \: sinA = m\: sinB \: \implies\sf \: \boxed{\sf \: m= \dfrac{sinA}{sinB} \: }$

Now, Consider

$\sf \: \dfrac{ {m}^{2} - 1}{ {n}^{2} - 1}$

On substituting the value of m and n we get

$\sf \: = \: \dfrac{ {\left(\dfrac{sinA}{sinB} \right)}^{2} - 1}{ {\left(\dfrac{tanA}{tanB} \right)}^{2} - 1}$

can be further rewritten as

$\sf \: = \: \dfrac{ {\left(\dfrac{sinA}{sinB} \right)}^{2} - 1}{ {\left(\dfrac{sinA \: cosB}{cosA \: sinB} \right)}^{2} - 1}$

$\sf \: = \: \dfrac{\dfrac{ {sin}^{2}A }{ {sin}^{2}B } - 1}{\dfrac{ {sin}^{2} A \: {cos}^{2} B}{ {cos}^{2} A \: {sin}^{2} B} - 1}$

$\sf \: = \: \dfrac{\dfrac{ {sin}^{2}A - {sin}^{2}B }{ {sin}^{2}B }}{\dfrac{ {sin}^{2} A \: {cos}^{2} B -{cos}^{2} A \: {sin}^{2} B }{ {cos}^{2} A \: {sin}^{2} B} }$

$\sf \: = \: \dfrac{{sin}^{2} A - {sin}^{2} B}{{sin}^{2} A \: {cos}^{2} B - {cos}^{2} A \: {sin}^{2} B} \times {cos}^{2}A$

$\sf \: = \: \dfrac{{sin}^{2} A - {sin}^{2} B}{{sin}^{2} A(1 - {sin}^{2} B) - (1 - {sin}^{2} A) {sin}^{2} B} \times {cos}^{2}A$

$\sf \: = \: \dfrac{{sin}^{2} A - {sin}^{2} B}{{sin}^{2} A - {sin}^{2}A{sin}^{2} B - {sin}^{2}B + {sin}^{2} A{sin}^{2} B} \times {cos}^{2}A$

$\sf \: = \: \dfrac{{sin}^{2} A - {sin}^{2} B}{{sin}^{2} A - {sin}^{2}B } \times {cos}^{2}A$

$\sf \: = \: {cos}^{2}A$

Hence,

$\implies\sf \: \dfrac{ {m}^{2} - 1}{ {n}^{2} - 1} = {cos}^{2}A$

$\rule{190pt}{2pt}$

Additional Information

$\begin{gathered}\: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sinx = \dfrac{1}{cosecx} }\\ \\ \bigstar \: \bf{cosx = \dfrac{1}{secx} }\\ \\ \bigstar \: \bf{tanx = \dfrac{sinx}{cosx} = \dfrac{1}{cotx} }\\ \\ \bigstar \: \bf{cot x= \dfrac{cosx}{sinx} = \dfrac{1}{tanx} }\\ \\ \bigstar \: \bf{cosec x = \dfrac{1}{sinx} }\\ \\ \bigstar \: \bf{secx = \dfrac{1}{cosx} }\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 } \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$