4 A = 3 A = 3/4 = B/P P = 4 , B = 3 Now, H^2 = P^2 + B^2 H^2 = 3^2 + 4^2 H^2 = 5^2 H = 5 cosec^2 A cot^2 A nbs ...

431

You visited us 431 times! Enjoying our articles? Unlock Full Access!

Circular Measurement of Angle

cosec2 A - co...

Question

$c o s e c^{2} A - c o t^{2} A$ .

Open in App

Solution

$4 cot A = 3$
$cot A = \frac{3}{4} = \frac{B}{P}$
$P = 4$ , $B = 3$
Now, $H^{2} = P^{2} + B^{2}$
$H^{2} = 3^{2} + 4^{2}$
$H^{2} = 5^{2}$
$H = 5$
$c o s e c^{2} A - c o t^{2} A = (\frac{H}{P})^{2} - (\frac{B}{P})^{2}$
= $(\frac{5}{4})^{2} - (\frac{3}{4})^{2}$
= $\frac{25}{16} - \frac{9}{16}$
= $\frac{16}{16}$
= $1$

Suggest Corrections

Similar questions

3 - c o t^{2} A + c o s e c^{2} A

Prove that

Cot²A × cosec²A - cot²B × cosec² B = cot²A - cot² B

In the given figure, $\frac{B C}{A C}$ = $\frac{24}{25}$ .Find the value of $c o s e c^{2} A$ – $c o t^{2} A$ .

\frac{c o s A - s i n A + 1}{c o s A + s i n - 1} = c o s e c A + c o t A,

c o s e c^{2} A = 1 + c o t^{2} A

\frac{(c o s A - s i n A + 1)}{(c o s A + s i n A - 1)} = c o s e c A + c o t A

c o s e c^{2} A = 1 + c o t^{2} A

Join BYJU'S Learning Program