Question

options for the answer is available ​

Answer 1

Answer:

180 - A or 180 - B or 180 - C belongs to 2nd quadrant and tan is negative in 2nd quadrant.

360 - B belongs to 4th quadrant and tan is negative in 4th quadrant.

540 - C belongs to 2nd quadrant and tan is negative in 2nd quadrant.

Answer 2

Question :

If A , B and C are the angles of a triangle , then find the value of

$\sf{ \dfrac{ \tan ( B + C ) + \tan ( C + A ) + \tan ( A + B )}{ \tan ( \pi - A ) + \tan ( 2 \pi - B ) + \tan ( 3 \pi - C )}}$

Answer

Given : -

If A , B and C are the angles of a triangle .

Required to find : -

• Value of

$\sf{ \dfrac{ \tan ( B + C ) + \tan ( C + A ) + \tan ( A + B )}{ \tan ( \pi - A ) + \tan ( 2 \pi - B ) + \tan ( 3 \pi - C )}}$

Solution : -

If A , B and C are the angles of a triangle .

We need to find the value of ;

$\sf{ \dfrac{ \tan ( B + C ) + \tan ( C + A ) + \tan ( A + B )}{ \tan ( \pi - A ) + \tan ( 2 \pi - B ) + \tan ( 3 \pi - C )}}$

Since,

We know that ;

In a traingle , sum of all its three angles is 180° .

However,

π radians = 180°

This implies ;

A + B + C = π

tan ( A + B ) = tan ( π - c )

tan ( π - c ) = - tan c

[ From the identity : tan ( x - A ) = - tan A ( x = theta ) ]

Similarly,

tan ( B + C ) = tan ( π - A )

tan ( π - A ) = - tan A

Similarly,

tan ( C + A ) = tan ( π - B )

tan ( π - B ) = - tan B

Hence,

• tan ( A + B ) = - tan C

• tan ( C + A ) = - tan B

• tan ( B + C ) = - tan A

However,

• tan ( π - A ) = - tan A

[ From the identity Identity ]

• tan ( 2π - B ) = - tan B

[ From the identity ; tan ( 2π - x ) = - tan x ]

• tan ( 3π - C ) = - tan C

[ From the identity ; tan ( 3π - x ) = - tan x ]

Now,

Let's find the value of ;

$\sf{ \dfrac{ \tan ( B + C ) + \tan ( C + A ) + \tan ( A + B )}{ \tan ( \pi - A ) + \tan ( 2 \pi - B ) + \tan ( 3 \pi - C )}}$

$\displaystyle \sf \dfrac{ \tan ( A + B ) + \tan ( B + C ) + \tan ( C + A )}{ - \tan A + ( - \tan B ) + ( - \tan C )} \\ \\ \\ \sf \dfrac{ \tan ( A + B ) + \tan ( B + C ) + \tan ( C + A )}{ - \tan A - \tan B - \tan C } \\ \\ \\ \tt Since , \\ \tt \bullet \tan ( A + B ) = - \tan C \\ \tt \bullet \tan ( B + C ) = - \tan A \\ \tt \bullet \tan ( C + A ) = - \tan B \\ \textsf{ Substituting these values we get ; }\\ \\ \\ \sf \dfrac{- \tan C + ( - \tan A ) + ( - \tan B )}{- \tan A - \tan B - \tan C} \\ \\ \\ \sf \dfrac{ - \tan C - \tan A - \tan B }{- \tan A - \tan B - \tan C} \\ \\ \\ \sf \dfrac{ - \tan A - \tan B - \tan C}{ - \tan A - \tan B - \tan C} \\ \\ \textsf{Since, the numerator and denominator are alike they get cancelled} \\ \\ \\ \red{ \implies{ 1}}$

Therefore,

$\sf{ \dfrac{ \tan ( B + C ) + \tan ( C + A ) + \tan ( A + B )}{ \tan ( \pi - A ) + \tan ( 2 \pi - B ) + \tan ( 3 \pi - C )} = 1 }$

1 is the correct option ✓

Additional Information :

Complementary Angles

sin x = cos ( 90° - x )

cos x = sin ( 90° - x )

tan x = cot ( 90° - x )

cot x = tan ( 90° - x )

sec x = cosec ( 90° - x )

cosec x = sec ( 90° - x )

Pythagorean Identities

sin² x + cos² x = 1

sec² x - tan² x = 1

cosec² x - cot² x = 1

Some Important formulae

sin ( A + B ) = sin A cos B + cos A sin B

cos ( A + B ) = cos A cos B - sin A sin B

tan ( A + B ) = tan A + tan B/1 - tan A tan B

tan ( A - B ) = tan A - tan B/1 + tan A tan B

cos ( A - B ) = cos A cos B - sin A sin B

sin ( A - B ) sin A cos B - cos A sin B

Reciprocals ratios

cosec x = 1/sin x

sec x = 1/cos x

cot x = 1/tan x

Trick to convert radians to degrees

• 1 radian = 180°/π

Trick to convert degrees to radians

• 1 degree = π/180°