Question

if sin theta = p/q then value of tan theta+sec theta is?

Answer 1

Step-by-step explanation:

You can do also by this method if you don't learnt formulas

Answer 2

Given:

sin θ = p/q

To find:

The value of tan θ + sec θ.

Solution:

As we know that in a right-angled triangle ABC where angle B = 90°, angle C = θ, AC = hypotenuse, BC = base and AB = height,

the sin θ is

$= \frac{height}{hypotenuse}$

$= \frac{AB}{AC}$

Also,

$cos \: θ = \frac{base}{hypotenuse} = \frac{BC}{AC}$

$tan \: θ = \frac{sin \:θ }{cos \: θ }$

$sec \: θ = \frac{1}{Cos \: θ }$

and,

From Pythagoras theorem,

${AC}^{2} = {AB}^{2} + {BC}^{2}$

Now, as given sin θ = p/q,

So,

$cos \: θ = \frac{\sqrt{ {q}^{2} - {p}^{2} } }{q}$

$sec \: θ = \frac{1}{ \frac{ \sqrt{ {q}^{2} - {p}^{2} } }{q} } = \frac{q}{ \sqrt{ {q}^{2} - {p}^{2} } }$

$tan \: θ = \frac{p}{ \sqrt{ {q}^{2} - {p}^{2} } }$

( using Pythagoras theorem)

Hence,

$tan \: θ + sec \: θ \: =$

$= \frac{p}{ \sqrt{ {q}^{2} - {p}^{2} } } + \frac{q}{ \sqrt{ {q}^{2} - {p}^{2} } }$

$= \frac{p \: + \: q}{ \sqrt{ {q}^{2} - {p}^{2} } }$

Hence, the value of tan θ + sec θ is

$= \frac{p \: + \: q}{ \sqrt{ {q}^{2} - {p}^{2} } }$