Question

What is the value of tanx+cotx if :

1)cosx=7/25
2)sinx=3/5
3)cosx=1/5

Give full detailed answer with steps .

Answer 1

1. cosx=b/h

b=7

h=25

p^2=h^2-b^2

p^2=25^2-7^2

p^2=625-49

p^2=576

p=24

so,tanx+cotx. p/b +b/p

24/7+7/24

583/168

2.sinx=p/h

p=3

h=5

b^2=h^2-p^2

b^2=5^2-3^2

b^2=25-9

b^2=16

b=4

so,tanx+cotx. p/b+b/p

3/5 +5/3

34/15

3.cosx=b/h

b=1

h=5

p^2=h^2-b^2

p^2=5^2-1^2

p^2=25-1

p^2=24

p=√24

now,tanx+cotx. p/b +b/p

√24/1 +1/√24

24+1/√24

Answer 2

Answer:

We are required to find tan x + cot x .

1) cos x = 7/25

We know that sin²x + cos²x = 1

⇒ sin²x + ( 7/25 )² = 1

⇒ sin²x + 49/625 = 1

⇒ sin²x = 1 - 49/625

⇒ sin²x = ( 625 - 49 ) / 625

⇒ sin²x = 576/625

⇒ sin x = √( 576/625 )

⇒ sin x = 24/25

We know that cot x = cos x / sin x

⇒ cot x = ( 7/25 ) / ( 24/25 )

⇒ cot x = 7/24

We know that tan x = 1/cot x

⇒ tan x = 24/7

tan x + cot x = 24/7 + 7/24

⇒ tan x + cot x = ( 24² + 7² ) / ( 24×7 )

⇒ tan x + cot x = ( 576 + 49 ) / 168

⇒ tan x + cot x = 625/168

2) sin x = 3/5

We know that sin²x + cos²x = 1

⇒ cos²x + ( 3/5 )² = 1

⇒ cos²x + 9/25 = 1

⇒ cos²x = 1 - 9/25

⇒ cos²x = ( 25 - 9 ) / 25

⇒ cos²x = 16/25

⇒ cos x = 4/5

We know that cot x = cos x / sin x

⇒ cot x = ( 4/5 ) / ( 3/5 )

⇒ cot x = 4/3

We know that tan x = 1/cot x

⇒ tan x = 3/4

tan x + cot x = 3/4 + 4/3

⇒ tan x + cot x = ( 3² + 4² ) / ( 4×3 )

⇒ tan x + cot x = ( 9 + 16 ) / 12

⇒ tan x + cot x = 25/12

3)

cos x = 1/5

We know that sin²x + cos²x = 1

⇒ sin²x + ( 1/5 )² = 1

⇒ sin²x + 1/25 = 1

⇒ sin²x = 1 - 1/25

⇒ sin²x = ( 25 - 1 ) / 625

⇒ sin²x =24/25

⇒ sin x = √( 24/25 )

⇒ sin x = √24/5

We know that cot x = cos x / sin x

⇒ cot x = ( 1/5 ) / ( √24/5 )

⇒ cot x = 1/√24

We know that tan x = 1/cot x

⇒ tan x = √24

tan x + cot x = √24 + 1/√24

⇒ tan x + cot x = ( 24 + 1 )/√24

⇒ tan x + cot x = 25/√24

⇒ tan x + cot x = 25√24/24