If cot theta =root 7 show that cosec ^2 theta -sec^2theta /cosec^2thet

1.	If cot theta =root 7 show that cosec ^2 theta -sec^2theta /cosec^2theta +sec^2theta =3/4As given in the picture
Answer» ANSWER: $\frac{(cosec^{2}\theta-sec^{2}\theta)}{(cosec^{2}\theta+sec^{2}\theta)}\\=\frac{3}{4}$ Step-by-step explanation: $Given \: cot\theta = \sqrt{7}--(<klux>1</klux>)$ $LHS =\frac{(cosec^{2}\theta-sec^{2}\theta)}{(cosec^{2}\theta+sec^{2}\theta)}$ $=\frac{\frac{1}{sin^{2}\theta}-\frac{1}{cos^{2}\theta}}{\frac{1}{sin^{2}\theta}+\frac{1}{cos^{1}\theta}}$ 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)*/ $=\frac{7-1}{7+1}\\=\frac{6}{8}\\=\frac{3}{4}\\=RHS$ •••♪

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

Answer»

ANSWER:

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

Step-by-step explanation:

$Given \: cot\theta = \sqrt{7}--(<klux>1</klux>)$

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

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

MULTIPLY NUMERATOR and denominator by $cos^{2}\theta$ ,we get

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