Question

If sec theta + tan theta = m , show that m^2-1/m^2+1 = sin theta

Answer 1

Hope It is help full

Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Questions}}}}$

Given,

secθ + tanθ = m

$\Large{\boxed{\sf\:{Squaring\;Both\;Sides}}}$

(secθ + tanθ)² = m²

sec²θ+ tan²θ + 2secθtan θ = m²

Now,

LHS

$\rm \: \: \: \: \: \: \: \: \:=\dfrac{m^2-1}{m^2+1}$

$\rm \: \: \: \: \: \: \: \: \:=\dfrac{sec^2\theta +tan^2\theta +2sec\theta tan\theta -1}{sec^2\theta +tan^2\theta +2sec\theta tan\theta +1}$

$\rm \: \: \: \: \: \: \: \: \:=\dfrac{(sec^2\theta -1)+tan^2\theta +2sec\theta tan\theta}{sec^2\theta +(1+tan^2\theta) +2sec\theta tan\theta}$

$\rm \: \: \: \: \: \: \: \: \:=\dfrac{tan^2\theta +tan^2\theta +2sec\theta tan\theta}{sec^2\theta +sec^2\theta +2sec\theta tan\theta}$

$\rm \: \: \: \: \: \: \: \: \:=\dfrac{sin\theta /cos\theta}{1/cos\theta}$

= sinθ

$\huge{\boxed{\sf\:{LHS=RHS}}}$

$\boxed{\begin{minipage}{11 cm} Fundamental Trignometric Indentities \\ \\ \sin^{2}\theta+\cos^{2}\theta =1 \\ \\ 1+tan^{2}\theta=\sec^{2}\theta \\ \\ 1 + cot^{2}\theta=\text{cosec}^2\theta \\ \\ tan\theta =\dfrac{sin\theta}{cos\theta} \\ \\ cot\theta =\dfrac{cos\theta}{sin\theta} \end{minipage}}$