Question

If cot B = $\frac{12}{5}$ prove that ${\tan }^{2}B-{\sin }^{2}B={\sin }^{4}B{\sec }^{2}B$

Answer 1

We have

Cot B = $\frac{base}{perpendicular}=\frac{12}{5}$

so, we draw a right angle ABC , right angle at C such that

Base = BC = 12 units and perpendicular = AC = 5

by pythagoras theorem , we have

$A{B}^2=B{C}^2+A{C}^2$

$\Rightarrow A{B}^2={12}^2+5^2$ = 169

$\Rightarrow AB=\sqrt{169}$ = 13

$\therefore sinB=\frac{AC}{AB}=\frac{5}{13},tanB=\frac{AC}{BC}=\frac{5}{12},secB=\frac{AB}{BC}=\frac{13}{12}$

Now , LHS = $ta{n}^{2}B-si{n}^{2}B$

$\Rightarrow LHS=(tanB{)}^2-(sinB{)}^2$

$\Rightarrow LHS={\left(\frac{5}{12}\right)}^2-{\left(\frac{5}{13}\right)}^2=\frac{25}{144}-\frac{25}{169}$

$\Rightarrow LHS=25(\frac{1}{144}-\frac{1}{169})=25\left(\frac{169-144}{144\times 169}\right)$

$\Rightarrow LHS=25\times \frac{25}{144\times 169}=\frac{25\times 25}{144\times 169}$

$\Rightarrow LHS=\frac{5^2\times 5^2}{{12}^2\times {13}^3}$

and RHS = $si{n}^{4}Bse{c}^{2}B$

$\Rightarrow =\left(sinB{)}^4\right(secB{)}^2$

$\Rightarrow ={\left(\frac{5}{13}\right)}^4\times {\left(\frac{13}{14}\right)}^2=\frac{5^4\times {13}^2}{{13}^4\times {12}^2}$

$\Rightarrow =\frac{5^4}{{13}^2\times {12}^2}=\frac{5^2\times 5^2}{{13}^2\times {12}^2}$

From (i) and (ii) , we have

$ta{n}^{2}B-si{n}^{2}B=si{n}^{4}Bse{c}^{2}B$