Question

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

Answer 1

Answer:

Hope it helps ....

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Answer 2

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

Given,

sinθ + cosθ = m

secθ + cosecθ = n

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

(sinθ + cosθ)² = m²

sin²θ + cos²θ + 2sinθcosθ = m²

2sinθcosθ = m² - 1 ............ (1)

And,

secθ + cosecθ = n

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

$\rm \: \: \: \: \: \: \: \: \:=\dfrac{sin\theta +cos\theta}{sin\theta cos\theta}=n$

$\rm \: \: \: \: \: \: \: \: \:=\dfrac{m}{sin\theta cos\theta}=n$

$\rm \: \: \: \: \: \: \: \: \: sin\theta cos\theta=\dfrac{m}{n}..... (2)$

From (1) and (2)

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

2m = n(m² - 1)