Question

Hlo everyone




using trigonometric relation no finding base or perpendicular


no spamಥ‿ಥ
$\cos \theta = \sqrt{2}$
Then find cot thetha and sec theta + Tan theta ​

Answer 1

Question :

If cos θ = √2 then find :

• cot θ

• sec θ + tan θ

Solution :

1) cot θ

To get cot θ we need to find sin θ

Using square relations we get :

sin² θ + cos² θ = 1

sin² θ = 1 - cos² θ

Putting cos θ value we get

sin² θ = 1 - (√2)²

sin² θ = 1 - 2

sin² θ = - 1

sin θ = √- 1

sin θ = 1

So we got sin θ as 1

Now, to get cot θ

To get cot θ we need to apply the formula : cot θ = cos θ/sin θ

Putting the values we get

cot θ = √2/1

cot θ = √2

So the answer is √2

ii) sec θ + tan θ

We know that

sec θ is the reciprocal of cos θ

tan θ is the reciprocal of cot θ

So,

sec θ = 1/ cos θ

• sec θ = 1/√2

tan θ = 1/cot θ

• tan θ = 1/√2

Now, according to the question we need to add sec θ and tan θ

sec θ + tan θ

$\sf \frac{1}{ \sqrt{2} } + \frac{1}{ \sqrt{2} }$

$\frac{2}{2 \sqrt{2} }$

$\frac{ \cancel{2}}{ \cancel{2} \sqrt{2} }$

$\frac{1}{ \sqrt{2} }$

So 1/√2 is the answer

KNOW MORE :

Square Relations :

• sin² θ + cos² θ = 1

• sec² θ – tan² θ = 1

• cosec² θ – cot² θ= 1

Quotient Relations :

• sin θ× cosec θ = 1

• cos θ × sec θ = 1

• tan θ × cot θ = 1

Basic :

sin ∅ = P/H

cos ∅ = B/H

tan ∅ = P/B

cot = B/P

sec = H/B

cosec = H/P

Here,

• P refers Perpendicular or Height

• B refers Base

• H refers Hypotentuse

Regards

# BeBrainly

Answer 2

Answer:

If cos θ = √2 then find :

cot θ

sec θ + tan θ

Solution :

1) cot θ

To get cot θ we need to find sin θ

Using square relations we get :

sin² θ + cos² θ = 1

sin² θ = 1 - cos² θ

Putting cos θ value we get

sin² θ = 1 - (√2)²

sin² θ = 1 - 2

sin² θ = - 1

sin θ = √- 1

sin θ = 1

So we got sin θ as 1

Now, to get cot θ

To get cot θ we need to apply the formula : cot θ = cos θ/sin θ

Putting the values we get

cot θ = √2/1

cot θ = √2

So the answer is √2

ii) sec θ + tan θ

We know that

sec θ is the reciprocal of cos θ

tan θ is the reciprocal of cot θ

So,

sec θ = 1/ cos θ

sec θ = 1/√2

tan θ = 1/cot θ

tan θ = 1/√2

Now, according to the question we need to add sec θ and tan θ

sec θ + tan θ

\sf \frac{1}{ \sqrt{2} } + \frac{1}{ \sqrt{2} }

2

1

+

2

1

\frac{2}{2 \sqrt{2} }

2

2

2

\frac{ \cancel{2}}{ \cancel{2} \sqrt{2} }

2

2

2

\frac{1}{ \sqrt{2} }

2

1

So 1/√2 is the answer

KNOW MORE :

Square Relations :

sin² θ + cos² θ = 1

sec² θ – tan² θ = 1

cosec² θ – cot² θ= 1

Quotient Relations :

sin θ× cosec θ = 1

cos θ × sec θ = 1

tan θ × cot θ = 1

Basic :

sin ∅ = P/H

cos ∅ = B/H

tan ∅ = P/B

cot = B/P

sec = H/B

cosec = H/P

Here,

P refers Perpendicular or Height

B refers Base

H refers Hypotentuse