First slide

Compound angles

Question

lf A, B, C are the angles of a given triangle ABC. If cosA cosB , $\cos C = \frac{\sqrt{3} - 1}{8}$ and sinA sinB $\sin C = \frac{3 + \sqrt{3}}{8}$ ,then

Moderate

Question

The value of tanA+ tanB +tanC is

A

$\frac{3 + \sqrt{3}}{\sqrt{3} - 1}$
B

$\frac{\sqrt{3} + 4}{\sqrt{3} - 1}$
C

$\frac{6 - \sqrt{3}}{\sqrt{3} - 1}$
D

$\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - 1}$

Solution

$\cos Acos Bcos C = \frac{\sqrt{3} - 1}{8}, \sin Asin Bsin C = \frac{3 + \sqrt{3}}{8}$
$\Rightarrow \tan Atan Btan C = \frac{3 + \sqrt{3}}{\sqrt{3} - 1} - - - - (i)$
For a triangle,
$\tan A + \tan B + \tan C = \tan Atan Btan C = \frac{3 + \sqrt{3}}{\sqrt{3} - 1} - - - - (ii)$
Now, $A + B + C = π \Rightarrow \cos (A + B + C) = \cos π$
$\begin{aligned} \Rightarrow \cos Acos Bcos C (1 - \tan Atan B - \tan Btan C - \tan Ctan A) \\ \Rightarrow \tan Atan B + \tan Btan C + \tan Ctan A = 5 + 4 \sqrt{3} - - - - (iii) \end{aligned}$
From(i ), (ii) and (iii)
tanA, tanB, tanC are the roots of the equation
$x^{3} - (\frac{3 + \sqrt{3}}{\sqrt{3} - 1}) x^{2} + (5 + 4 \sqrt{3}) x - (3 + 2 \sqrt{3}) = 0 - - - - (iv)$
x = 1 is the root of above equation so,
$\begin{aligned} (x - 1) {x^{2} - (2 + 2 \sqrt{3}) x + (3 + 2 \sqrt{3})} = 0 \\ (x - 1) (x - \sqrt{3}) (x - (2 + \sqrt{3})) = 0 \\ x = 1, \sqrt{3}, 2 + \sqrt{3} \end{aligned}$
$\tan A = 1, \tan B = \sqrt{3}, \tan C = 2 + \sqrt{3}$
Therefore, $A = 45^{\circ}, B = 60^{\circ}, C = 75^{\circ}$

Question

The value of tanA tanB tanB tanC+tanC tanA is

A

$5 - 4 \sqrt{3}$
B

$5 + 4 \sqrt{3}$
C

$6 + \sqrt{3}$
D

$6 - \sqrt{3}$

Solution

$\cos Acos Bcos C = \frac{\sqrt{3} - 1}{8}, \sin Asin Bsin C = \frac{3 + \sqrt{3}}{8}$
$\Rightarrow \tan Atan Btan C = \frac{3 + \sqrt{3}}{\sqrt{3} - 1} - - - - (i)$
For a triangle,
$\tan A + \tan B + \tan C = \tan Atan Btan C = \frac{3 + \sqrt{3}}{\sqrt{3} - 1} - - - - (ii)$
Now, $A + B + C = π \Rightarrow \cos (A + B + C) = \cos π$
$\begin{aligned} \Rightarrow \cos Acos Bcos C (1 - \tan Atan B - \tan Btan C - \tan Ctan A) \\ \Rightarrow \tan Atan B + \tan Btan C + \tan Ctan A = 5 + 4 \sqrt{3} - - - - (iii) \end{aligned}$
From(i ), (ii) and (iii)
tanA, tanB, tanC are the roots of the equation
$x^{3} - (\frac{3 + \sqrt{3}}{\sqrt{3} - 1}) x^{2} + (5 + 4 \sqrt{3}) x - (3 + 2 \sqrt{3}) = 0 - - - - (iv)$
x = 1 is the root of above equation so,
$\begin{aligned} (x - 1) {x^{2} - (2 + 2 \sqrt{3}) x + (3 + 2 \sqrt{3})} = 0 \\ (x - 1) (x - \sqrt{3}) (x - (2 + \sqrt{3})) = 0 \\ x = 1, \sqrt{3}, 2 + \sqrt{3} \end{aligned}$
$\tan A = 1, \tan B = \sqrt{3}, \tan C = 2 + \sqrt{3}$
Therefore, $A = 45^{\circ}, B = 60^{\circ}, C = 75^{\circ}$

Question

The value of tanA, tanB and tanC are

A

$1, \sqrt{3}, \sqrt{2}$
B

$1, \sqrt{3}, 2$
C

$1, 2, \sqrt{3}$
D

$1, \sqrt{3}, 2 + \sqrt{3}$

Solution

$\cos Acos Bcos C = \frac{\sqrt{3} - 1}{8}, \sin Asin Bsin C = \frac{3 + \sqrt{3}}{8}$
$\Rightarrow \tan Atan Btan C = \frac{3 + \sqrt{3}}{\sqrt{3} - 1} - - - - (i)$
For a triangle,
$\tan A + \tan B + \tan C = \tan Atan Btan C = \frac{3 + \sqrt{3}}{\sqrt{3} - 1} - - - - (ii)$
Now, $A + B + C = π \Rightarrow \cos (A + B + C) = \cos π$
$\begin{aligned} \Rightarrow \cos Acos Bcos C (1 - \tan Atan B - \tan Btan C - \tan Ctan A) \\ \Rightarrow \tan Atan B + \tan Btan C + \tan Ctan A = 5 + 4 \sqrt{3} - - - - (iii) \end{aligned}$
From(i ), (ii) and (iii)
tanA, tanB, tanC are the roots of the equation
$x^{3} - (\frac{3 + \sqrt{3}}{\sqrt{3} - 1}) x^{2} + (5 + 4 \sqrt{3}) x - (3 + 2 \sqrt{3}) = 0 - - - - (iv)$
x = 1 is the root of above equation so,
$\begin{aligned} (x - 1) {x^{2} - (2 + 2 \sqrt{3}) x + (3 + 2 \sqrt{3})} = 0 \\ (x - 1) (x - \sqrt{3}) (x - (2 + \sqrt{3})) = 0 \\ x = 1, \sqrt{3}, 2 + \sqrt{3} \end{aligned}$
$\tan A = 1, \tan B = \sqrt{3}, \tan C = 2 + \sqrt{3}$
Therefore, $A = 45^{\circ}, B = 60^{\circ}, C = 75^{\circ}$

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