Question

Express the trigonometric ratios sinA, secA and tanA in terms of cotA.​

Answer 1

Answer:

1) sinA= 1/cosecA

sinA = 1 / √(1+cot²A)

[ cot²A+ 1 = cosec²A,

cosecA= √( 1+cot²A)]

2) tanA= 1/cotA

3) secA= √(1+tan²A)

[sec²A= 1+tan²A , secA= √ (1+tan²A)]

secA= √(1+ (1/cot²A)) = √ (1+1/ cot²A)

secA = √(cot²A+1/cot²A)

secA= √1+cot²A/ (cotA)

HOPE IT HELPS!

Answer 2

$\large\sf{{ \cosec }^{2} A = 1 + {cot}^{2} A}$

$\large\sf{\frac{1}{ {cosec }^{2} A} = \frac{1}{1 + {cot}^{2} A}}$

$\large\sf{ { \sin }^{2} A = \frac{1}{1 + { \cot}^{2}A }}$

$\large\sf{sin \: A = \frac{1}{1 + {cot}^{2}A }}$

$\therefore$$\large\sf{sin \: a = \frac{1}{ \sqrt{1 + {cot}^{2} A} }}$

$\huge\sf\red{We\:know\:that,}$

$\large\sf{tanA=\frac{sinA}{cosA}}$

$\huge\sf\green{However,}$

$\large\sf{cotA=\frac{cosA}{sinA}}$

$\therefore$$\large\sf{tanA=\frac{1}{cotA}}$

$\huge\sf\purple{Also,}$

$\longrightarrow$$\large\sf{{ \sec }^{2} A = 1 + { \tan }^{2} A}$

$\longrightarrow$$\large\sf{1+\frac{1}{{cot}^{2}A}}$

$\longrightarrow$$\large\sf{\frac{ { \cot }^{2}A + 1 }{ { \cot }^{2} A}}$

$\longrightarrow$$\large\sf{\sec \: A = \frac{ \sqrt{ {cot}^{2} A + 1} }{cot \: A}}$