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Find the general solution of the equationSin5 x – cos5 x = 1/cosx – 1/sinx (sinx ≠ cosx )

Nirabhra Agrawal , 13 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:Hello student, please find answer to your question
$sin^{5}x - cos^{5}x = \frac{1}{cosx}-\frac{1}{sinx}$
$(sinx - cosx)(sin^{4}x+sin^{3}x.cosx+cos^{3}x.sinx+sin^{2}x.cos^{2}x+cos^{4}x) = \frac{sinx-cosx}{sinx.cosx}$ $sinx \neq cosx$
$sinx.cosx(sin^{4}x+sinx.cosx(sin^{2}x+cos^{2}x)+sin^{2}x.cos^{2}x+cos^{4}x) = 1$ $sinx.cosx(sin^{4}x+sinx.cosx+sin^{2}x.cos^{2}x+cos^{4}x) = 1$
$sinx.cosx(sin^{4}x+cos^{4}x+sinx.cosx+sin^{2}x.cos^{2}x) = 1$
$sin^{4}x+cos^{4}x = (sin^{2}x-\sqrt{2}sinx.cosx+cos^{2}x)(sin^{2}x+\sqrt{2}sinx.cosx+cos^{2}x)$ $sin^{4}x+cos^{4}x = (1-\sqrt{2}sinx.cosx)(1+\sqrt{2}sinx.cosx)$
$sin^{4}x+cos^{4}x = 1-2sin^{2}x.cos^{2}x$
$sinx.cosx(sin^{4}x+cos^{4}x+sinx.cosx+sin^{2}x.cos^{2}x) = 1$
$sinx.cosx(1-2sin^{2}x.cos^{2}x+sinx.cosx+sin^{2}x.cos^{2}x) = 1$
$sinx.cosx(1-sin^{2}x.cos^{2}x+sinx.cosx) = 1$
$\Rightarrow {2\pi n-2tan^{-1}(1-\sqrt{2}), 2\pi n-2tan^{-1}(1+\sqrt{2})}$