Question

If sin theta=(3)/(5) then find the value of cot^(2)theta+sec^(2)theta

Answer 1

Answer:

draw a right angle triangle

opposite side=3 , adjacent side=x and hypotenuse =5

H²=S²+S²

5²=3²+S²

25=9+S²

25-9=S²

16=S²

√16=S

S=4

adjacent side =4

cot theta=adj/opp=4/3

sec theta =hyp/adj=5/4

(4/3)²+(5/4)²

16/9+25/16

256+225/144

=481/144

Answer 2

Given :-

$\sf sin \theta = \dfrac{3}{5}$

To Find :-

$\sf cot^2 \theta +sec^2 \theta$

Solution :-

We know :-

$\bf cosec \theta = \dfrac{1}{sin \theta }$

$\sf\therefore cosec \theta = \dfrac{5}{3}$

$\bullet \bf \ cot^2\theta = cosec^2 \theta - 1$

$\longrightarrow \sf cot^2 \theta = \left(\dfrac{5}{3} \right)^2- 1 \\\\\longrightarrow cot^2 \theta = \dfrac{25}{9} - 1 \\\\\longrightarrow cot^2 \theta = \dfrac{25-9}{9} \\\\\longrightarrow cot^2 \theta = \dfrac{16}{9} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ..................(1)$

$\sf tan^2 \theta = \dfrac{1}{cot^2 \theta}$

$\therefore \sf tan^2 \theta = \dfrac{9}{16}$

$\bullet \ \bf sec^2 \theta= 1+tan^2 \theta$

$\longrightarrow \sf sec^2 \theta = 1+ \dfrac{9}{16} \\\\\longrightarrow sec^2 \theta = \dfrac{16+9}{16} \\\\\longrightarrow sec^2 \theta = \dfrac{25}{16} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...................(2)$

Adding eq(1) and eq(2), we get :-

$\longrightarrow \sf cot^2 \theta + sec ^2 \theta = \dfrac{16}{9} + \dfrac{25}{16} \\\\\longrightarrow cot^2 \theta +sec^2 \theta = \dfrac{(16 \times 16)+( 25 \times 9 )}{9 \times 16} \\\\\longrightarrow cot^2 \theta +sec^2 \theta = \dfrac{256+225}{144} \\\\\longrightarrow \bf cot^2 \theta +sec^2 \theta = \dfrac{481}{144}$