Question

tan A =1/√3 find the value of sec A cosec A + cosec A sec A​

Answer 1

Answer:

8√3/√3

Step-by-step explanation:

$tanA=\frac{1}{\sqrt{3} } =\frac{P}B}$

Perpendicular(P) = 1

Base(B) = √3

Applying Pythagoras theorem,

(H)² = (P)² + (B)²

(H)² = (1)² + (√3)²

(H)² = 1 + 3

(H)² = 4

H = √4

H = 2

Hypotenuse(H) = 2

$SecA = \frac{H}{B} =\frac{2}{\sqrt{3} } \\\\CosecA = \frac{H}{P} =\frac{2}{1}$

Sec A Cosec A + CosecA SecA​

$(\frac{2}{\sqrt{3} } )(\frac{2}{1} )+(\frac{2}{1} )(\frac{2}{\sqrt{3} } )\\\\\frac{4}{\sqrt{3} } +\frac{4}{\sqrt{3} } \\\\\frac{8}{\sqrt{3} } \\\\\frac{8}{\sqrt{3} } *\frac{\sqrt{3} }{\sqrt{3} } \\\\\frac{8\sqrt{3} }{3}$