Question

Tan (a) + tan(60 + a) - tan(60-a) is equal to i) 3tan3a ii) cot3a iii)sin3a iv)cot3a

Answer 1

Answer:

3 tan (a)

Step-by-step explanation:

$\tan(a) + \tan(60 + a) - \tan(60 - a)$

$\tan(60 + a) = \frac{ \tan(a) + \tan(60) }{1 - \tan(a) \tan(60) }$

$= \frac{ \tan(a) + \sqrt{3} }{1 - \tan(a) \sqrt{3} }$

$\tan(60 - a) = \frac{ \sqrt{3} - \tan(a) }{1 + \sqrt{3} \tan(a) }$

$\tan(a) + \tan(60 + a) - \tan(60 - a)$

$= \tan(a) + \frac{ \tan(a) + \sqrt{3} }{1 - \tan(a) \sqrt{3} } - \frac{ \tan(a) - \sqrt{3} }{1 + \sqrt{3} \tan(a) }$

Refer the attachment for the LCM step,

Next,

$= \tan( a ) + \frac{8 \tan(a) }{1 - 3 { \tan(a) }^{2} }$

$= \frac{9 \tan(a) - 3 { \tan(a) }^{3} }{1 - 3 { \tan(a) }^{2} }$

$= 3 \frac{3 \tan(a) - { \tan(a) }^{3} }{1 - { \tan(a) }^{2} }$

By using Trigonometric Identity,

$= 3 \tan(3a)$

Answer 2

Answer:

let tan(a)=t

tan3a=t3

tan(a+b)=(tana+tanb)÷(1-tanatanb)

tan60=1.7(root three value approx)

• t+ (1.7 + t) ÷(1- 1.7t) - ((1.7 - t)÷(1+ 1.7t))
• t + {(8t) ÷(1-3t²) }
• by lcm
• (9t-3t³)÷(1-3t²)
• tan3a=(3tana -tan³a)÷(1 -3tan²a)
• so 1-3tan²a=(3tana -tan³a)÷(3tana) substitute in above equation
• (9t-3t³)÷{(3t - t³) ÷t3}
• {(9t-3t³)÷{(3t - t³)}× t3
• {3}×t3
• 3t3
• 3tan3a