Question

If sin A=1upon3,evaluate cos A cosec A+tan A sac A

Answer 1

Answer:

Given that ;

$\rm \sin(A) = \dfrac{1}{3} = \dfrac{perpendicular}{hypotenuse}$

So, here we have -

• Perpendicular = 1
• Hypotenuse = 3

We need to find base. We can use Pythagoras theorem here.

$\rm Base {}^{2} = h {}^{2} - p {}^{2} \\ \\ \rm Base {}^{2} = (3) {}^{2} - (1) {}^{2} \\ \\ \rm Base {}^{2} = 9 - 1 \\ \\ \rm Base = \sqrt{8} = 2 \sqrt{2}$

$\rule{300}{2}$

$\rm \cos(A) = \frac{b}{h} = \frac{2 \sqrt{2} }{3} \\ \\ \rm \cosec(A) = \frac{h}{p} = \frac{3}{1} = 3$

And,

$\rm \tan(A) = \frac{p}{b} = \frac{1 }{2 \sqrt{2} } \\ \\ \rm \sec(A) = \frac{h}{b} = \frac{3}{2 \sqrt{2 } }$

Thus,

$\rm \cos(A) \cosec(A) + \tan(A) \sec(A) \\ \\ \frac{2 \sqrt{2} }{3 } \times 3 + \frac{1}{2 \sqrt{2} } \times \frac{3}{2 \sqrt{2} } \\ \\ \implies 2 \sqrt{2} + \frac{3}{8} \\ \\ \red{\boxed{ \therefore \bf \: \frac{16 \sqrt{2} + 3 }{8}}}$

Answer 2

Answer =>

$\mathfrak{\huge{\underline{\boxed{\red{\frac{16 \sqrt{2} + 3}{8}}}}}}$

Explanation =>

Given :-

$Sin (A) = \frac{1}{3} = \frac{Perpendicular}{hypotenuse}$

So,

Perpendicular = 1

Hypotenuse = 3

____________________________

We have to find base by Pythagoras theorem :-

(H)² = (B)² + (P)²

____________________________

So,

B² = H² - P²

______________[Put values]

B² = (3)² - (1)²

B² = 9-1

B² = 8

B = ±√8

B = ±√2×2×2

B = 2√2

____________________________

___________________

_________

We know that :-

• Cos(A) = b/h => 2√2/3
• Cosec(A) = h/p => 3
• Tan(A) = p/b =>1/2√2
• Sec(A)= h/b => 3/2√2

A.T.Q

Cos(A) * Cosec(A) + Tan(A)*Sec(A)

__________________________

______________________

__Put Values _______

2√2/3 * 3 + 1/2√2 * 3/2√2

» 2√2 + 3/8

» 16√2 + 3 / 8

____________[Answer]

Remember :-

• Sin A = p/h
• Cos A = b/h
• Tan A = p/b

And

• Cosec A = h/p
• Sec A = h/b
• Cot A = b/p