Exact Value of tan 142½°

How to find the exact value of tan 142½° using the value of sin 15° and cos 15°?

Solution:

For all values of the angle A and B we know that, tan (A + B) = $\frac{tan A + tan B}{1 - tan A tan B}$, 

sin A = 2 sin $\frac{A}{2}$ cos $\frac{A}{2}$ 

and

cos A = cos$^{2}$ $\frac{A}{2}$ – sin$^{2}$ $\frac{A}{2}$

Now tan 142½°

= tan (90 + 52½°)

= - cot 52½°

= $\frac{-1}{tan 52½°}$

= $\frac{-1}{tan (45° + 7½°}$

= - $\frac{1 - tan  7½°}{1 + tan  7½°}$

= - $\frac{cos  7½° - sin 7½°}{cos  7½° + sin 7½°}$

=  - $\frac{(cos  7½° - sin 7½°)(cos  7½° - sin 7½°)}{(cos  7½° + sin 7½°)(cos  7½° - sin 7½°)}$

= - $\frac{(cos  7½° - sin 7½°)^{2}}{cos^{2}  7½° - sin^{2} 7½°}$

= - $\frac{1 - 2 sin  7½°  cos 7½°}{cos^{2}  7½° - sin^{2} 7½°}$

= - $\frac{1 - sin 15°}{cos 15°}$

= - $\frac{1 - sin (45° - 30°)}{cos (45° - 30°)}$

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

= - $\frac{2√2 - √3 + 1}{√3 + 1}$

= - $\frac{(2√2 - √3 + 1)}{(√3 + 1)}$  $\frac{(√3 - 1)}{(√3 - 1)}$

= - $\frac{(2√2 - √3 + 1)(√3 - 1)}{3 - 1}$

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

= -[√2(√3 - 1) – (2 - √3)]

= -√6 + √2 + 2 - √3

= 2 + √2 - √3 - √6

● Submultiple Angles

• Trigonometric Ratios of Angle $\frac{A}{2}$
• Trigonometric Ratios of Angle $\frac{A}{3}$
• Trigonometric Ratios of Angle $\frac{A}{2}$ in Terms of cos A
• tan $\frac{A}{2}$ in Terms of tan A
• Exact value of sin 7½°
• Exact value of cos 7½°
• Exact value of tan 7½°
• Exact Value of cot 7½°
• Exact Value of tan 11¼°
  
• Exact Value of sin 15°
• Exact Value of cos 15°
• Exact Value of tan 15°
  
• Exact Value of sin 18°
• Exact Value of cos 18°
• Exact Value of sin 22½°
• Exact Value of cos 22½°
• Exact Value of tan 22½°
• Exact Value of sin 27°
• Exact Value of cos 27°
• Exact Value of tan 27°
  
• Exact Value of sin 36°
• Exact Value of cos 36°
• Exact Value of sin 54°
• Exact Value of cos 54°
• Exact Value of tan 54°
• Exact Value of sin 72°
• Exact Value of cos 72°
• Exact Value of tan 72°
• Exact Value of tan 142½°
  
• Submultiple Angle Formulae
  
• Problems on Submultiple Angles

11 and 12 Grade Math

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