Question

if tan alpha = p/q and tan beta = q/p ,then the angle (alpha + beta ) is equal to​

Answer 1

Answer:

please mark as brainliest please request

Answer 2

The value of $( \alpha + \beta )$ is $\bf \frac{\pi}{2}$.

Option 'B' is correct.

Given:

• $\tan( \alpha ) = \frac{p}{q}$
• $\tan( \beta ) = \frac{q}{p}$

To find:

• Find the angle $( \alpha + \beta )$
• A) π/3
• B) π/2
• C) π/6
• D) None of these

Solution:

Formula to be used:

• $\boxed{ \tan( \alpha + \beta ) = \frac{ \tan( \alpha ) + \tan( \beta ) }{1 - \tan( \alpha ) \tan( \beta ) }}$

Step 1:

Apply the formula .

As it is given

$\tan( \alpha ) = \frac{p}{q}$

and

$\tan( \beta ) = \frac{q}{p}$

So,

$\bf \tan( \alpha + \beta ) = \frac{ \frac{p}{q} + \frac{q}{p} }{1 - \frac{p}{q}. \frac{q}{p} }$

Step 2:

Simplify the expression in RHS.

$\tan( \alpha + \beta ) = \frac{ \frac{p ^{2} + {q}^{2} }{qp} }{1 - \frac{ \cancel p}{ \cancel q}. \frac{ \cancel q}{\cancel p} }$

$\tan( \alpha + \beta ) = \frac{ \frac{p ^{2} + {q}^{2} }{qp} }{1 - 1}$

$\tan( \alpha + \beta ) = \frac{ \frac{p ^{2} + {q}^{2} }{qp} }{0}$

As, something upon zero is a term which is not defined.

So,

$\tan( \alpha + \beta ) = \infty$

$( \alpha + \beta ) = {tan}^{ - 1} ( \infty)$

As value of tan(π/2) is ∞, or we can say not defined.

So,

$( \alpha + \beta ) = tan^{-1}\tan\left( \frac{\pi}{2} \right)$

$\bf ( \alpha + \beta ) = \frac{\pi}{2}$

Thus,

The value of $( \alpha + \beta )$ is $\bf \frac{\pi}{2}$.

Option 'B' is correct.

Learn more:

1) tan x + sec x = √3 find the value of x

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2) If tan theta =4/3 find the value of 2sin theta-3cos theta/2sin theta+3cos theta

https://brainly.in/question/6802000

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