Question

If sin theta + cos theta = m and sec theta + cosec theta = n, show that n(m^2 - 1) = 2m

Answer 1

Step-by-step explanation:

hope this will help you

Answer 2

AnswEr :

$\normalsize\:\bullet\:\sf\ sec\theta + cosec\theta = n$

$\underline{\bigstar\:\textsf{According \: to \: the \: question \: now:}}$

$\scriptsize\tt{\qquad\dag\ cosec\theta = \frac{1}{sin\theta}}$

$\scriptsize\tt{\qquad\dag\ sec\theta = \frac{1}{cos\theta}}$

⋆Taking left hand side :

$\normalsize\ : \implies\sf\ n(m^2 - 1)$

$\normalsize\ : \implies\sf\ \left(\frac{1}{cos\theta} + \frac{1}{sin\theta} \right) \left[(sin\theta + cos\theta)^{2} - 1) \right]$

$\normalsize\ : \implies\sf\ \left(\frac{sin\theta + cos\theta}{cos\theta sin\theta} \right) \left[sin^{2}\theta + cos^{2}\theta + 2sin\theta cos\theta - 1 \right]$

$\scriptsize\tt{\qquad\dag\ sin^2\theta + cos^2\theta = 1 }$

$\normalsize\ : \implies\sf\ \left(\frac{sin\theta + cos\theta}{cos\theta sin\theta} \right) \left[\cancel{1} + 2sin\theta cos\theta - \cancel{1} \right]$

$\normalsize\ : \implies\sf\ \left(\frac{sin\theta + cos\theta}{\cancel{cos\theta sin\theta}} \right) \left[2 \cancel{sin\theta cos\theta} \right]$

$\normalsize\ : \implies\sf\ 2(sin\theta + cos\theta) \quad\ ---(eq.1)$

⋆Taking right hand side :

$\normalsize\ : \implies\sf\ 2m$

$\normalsize\ : \implies\sf\ 2(sin\theta + cos\theta) \quad\ ---(eq.2)$

$\underline{\bigstar\:\textsf{from \: equation \: 1 \: and \: 2}}$

$\normalsize\ : \implies\sf\ 2(sin\theta + cos\theta) = 2(sin\theta + cos\theta)$

$\normalsize\ : \implies\sf\ L.H.S = R.H.S$

$\rule{170}1$

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\cos^2\theta=1-\sin^2\theta \\ \\ 4)1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5) \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\sec^2\theta=1+\tan^2\theta \\ \\ 8)\sec^2\theta-\tan^2\thetha=1 \\ \\ 9)\tan^2\theta=\sec^2\theta-1 \end{minipage}}$