Question

Given sinA=3/4,find cosA, tanA, cotA and secA​

Answer 1

Answer:

$given \: \: \: \: \: sinA = \frac{3}{4 } \\ \therefore \: cosA = \sqrt{1 - {sin}^{2}A } \\ \\ = \sqrt{1 - { (\frac{3}{4} }^{2}) } \\ \\ = \sqrt{1 - \frac{9}{16} } \\ \\ = \sqrt{ \frac{16 - 9}{16} } \\ \\ = \sqrt{ \frac{7}{16} } \\ \\ = \frac{ \sqrt{7} }{4} \\ \\ \\ tanA = \frac{sinA}{cosA} \\ \\ = \frac{ \frac{3}{4} }{ \frac{ \sqrt{7} }{4} } \\ \\ = \frac{3}{ \sqrt{7} } \\ \\ \\ cotA = \frac{1}{tanA} \\ \\ = \frac{1}{ \frac{3}{ \sqrt{7} } } \\ \\ = \frac{ \sqrt{7} }{3} \\ \\ \\ secA = \frac{1}{cosA} \\ \\ \frac{1}{ \frac{ \sqrt{7} }{4} } \\ \\ = \frac{4}{ \sqrt{7} } \\ \\ \\ cosecA = \frac{1}{sinA} \\ \\ = \frac{1}{ \frac{3}{4} } \\ \\ = \frac{4}{3}$

Answer 2

Answer:

givensinA=

4

3

∴cosA=

1−sin

2

A

=

1−(

4

3

2

)

=

1−

16

9

=

16

16−9

=

16

7

=

4

7

tanA=

cosA

sinA

=

4

7

4

3

=

7

3

cotA=

tanA

1

=

7

3

1

=

3

7

secA=

cosA

1

4

7

1

=

7

4

cosecA=

sinA

1

=

4

3

1

=

3

4