Question

if sin theta = 3 upon 4 prove that root cosec square theta - cot square theta upon sec square theta -1 =root 7 upon 3

Answer 1

here is your answer...

Answer 2

Solution:

$Given ,\\sin\theta=\frac{3}{4}---(1)$

$cos^{2}\theta \\= 1-sin^{2}\theta\\=1-\big(\frac{3}{4}\big)^{2}\\=1-\frac{9}{16}\\=\frac{16-9}{16}\\=\frac{7}{16}$

$cos\theta= \sqrt{\frac{7}{16}}\\=\frac{\sqrt{7}}{4}--(2)$

$Now, \\Value \: of \:\sqrt{ \frac{cosec^{2}\theta-cot^{2}\theta}{sec^{2}\theta-1}}$

$=\sqrt{\frac{1}{tan^{2}\theta}}$

$=\frac{1}{tan\theta}\\=\frac{1}{\frac{sin\theta}{cos\theta}}$

$=\frac{cos\theta}{sin\theta}\\=\frac{\frac{\sqrt{7}}{4}}{\frac{3}{4}}$

$=\frac{7}{4} \times \frac{3}{4}$

$=\frac{\sqrt{7}}{3}$

Therefore,

$Now, \\Value \: of \:\sqrt{ \frac{cosec^{2}\theta-cot^{2}\theta}{sec^{2}\theta-1}} = \frac{\sqrt{7}}{3}$

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