Math, asked by selvantamil9686, 5 months ago

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

Answers

Answered by kbheemashankar23
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

Answered by Ataraxia
24

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}

Similar questions