Question

[tanx+1/cosx]2 +[tanx-1/cosx]2=2[1+sin2x/1-sin2x]

Answer 1

Answer:

$\left(tanx +\frac{1}{cosx}\right)^{2}+\left(tanx -\frac{1}{cosx}\right)^{2}\\=2\left(\frac{1+sin^{2}x}{1-sin^{2}x}\right)$

Step-by-step explanation:

$LHS = \left(tanx +\frac{1}{cosx}\right)^{2}+\left(tanx -\frac{1}{cosx}\right)^{2}$

$= (tanx+secx)^{2}+(tanx-secx)^{2}$

$=2(tan^{2}x+sec^{2}x)$

$=2\left(\frac{sin^{2}x}{cos^{2}x}+\frac{1}{cos^{2}x}\right)$

$=2\left(\frac{sin^{2}x+1}{cos^{2}x}\right)$

$=2\left(\frac{1+sin^{2}x}{1-sin^{2}x}\right)$

$=RHS$

Therefore,

$\left(tanx +\frac{1}{cosx}\right)^{2}+\left(tanx -\frac{1}{cosx}\right)^{2}\\=2\left(\frac{1+sin^{2}x}{1-sin^{2}x}\right)$

•••♪