Question

If cot theta =root 7 show that cosec ^2 theta -sec^2theta /cosec^2theta +sec^2theta =3/4As given in the picture

Answer 1

Answer:

$\frac{(cosec^{2}\theta-sec^{2}\theta)}{(cosec^{2}\theta+sec^{2}\theta)}\\=\frac{3}{4}$

Step-by-step explanation:

$Given \: cot\theta = \sqrt{7}--(1)$

$LHS =\frac{(cosec^{2}\theta-sec^{2}\theta)}{(cosec^{2}\theta+sec^{2}\theta)}$

$=\frac{\frac{1}{sin^{2}\theta}-\frac{1}{cos^{2}\theta}}{\frac{1}{sin^{2}\theta}+\frac{1}{cos^{1}\theta}}$

Multiply numerator and denominator by $cos^{2}\theta$ ,we get

$=\frac{\frac{cos^{2}\theta}{sin^{2}\theta}-\frac{cos^{2}\theta}{cos^{2}\theta}}{\frac{cos^{2}\theta}{sin^{2}\theta}+\frac{cos^{2}\theta}{cos^{1}\theta}}$

$=\frac{cot^{2}\theta-1}{cot^{2}\theta +1}$

$=\frac{(\sqrt{7})^{2}-1}{(\sqrt{7})^{2}+1}$

/* From(1)*/

$=\frac{7-1}{7+1}\\=\frac{6}{8}\\=\frac{3}{4}\\=RHS$

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Answer 2

Answer:

3/4 is correct answer

Step-by-step explanation:

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