Question

If cot = root7. show that
cosec^2 -sec^2÷cosec^2+sec^2 =3÷4
​

Answer 1

Step-by-step explanation:

$\huge\underline{\overline{\mid{\bold{\blue{\mathcal{Question:-}}\mid}}}}$

$If \: cot \: x \: = \sqrt{7}$

Show that

$\huge\frac{ {cosec}^{2}x - {sec}^{2} x}{ {cosec}^{2} x + {sec}^{2} x} = \frac{3}{4}$

$\huge\underline{\overline{\mid{\bold{\blue{\mathcal{Hint:-}}\mid}}}}$

• We will prove the Left hand side to the Right hand side by finding the values of cosec x and sec x using cot x and the Pythagoras theorem.

$\huge\underline{\overline{\mid{\bold{\blue{\mathcal{Required Solution:-}}\mid}}}}$

☆ Given:-

Cot x = √7

$or \: \frac{base}{perpendicular} = \frac{ \sqrt{7} }{1}$

Using Pythagoras theorem in ∆ ABC [see the attachment]

we have;

${AC}^{2} = {AB}^{2} + {BC}^{2}$

$AC {}^{2} = 1 + 7 = > AC = \sqrt{8}$

$\mathcal{\green{Now,}}$

$\csc(x) = \frac{H}{P} = \frac{ \sqrt{8} }{1}$

$\sec(x) = \frac{H}{B} = \frac{ \sqrt{8} }{ \sqrt{7} }$

$\fbox{\purple{Using \ the \ values \ for \ the \ LHS}}$

$\mathcal{\green{Now,}}$

$\huge \frac{ \csc(x) ^{2} - {sec}^{2} x }{ { \csc(x) }^{2} + \sec(x ) ^{2} } = \frac{ { (\sqrt{8}) }^{2} - (\frac{ \sqrt{8} }{ \sqrt{7} } ) {}^{2} }{ { \sqrt{8} }^{2} + (\frac{ \sqrt{8} }{ \sqrt{7} } ) {}^{2} }$

$\huge = > \frac{8 - \frac{8}{7} }{8 + \frac{8}{7} }$

$\huge = > \frac{ \frac{56 - 8}{7} }{ \frac{56 + 8}{7} } = \frac{48}{64} \div \frac{16}{16} = \frac{3}{4}$

$\rightarrow\small\bf LHS=RHS$

□ Hence, proved.

$\mathfrak{\red{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ @MissTranquil}}$