Question

find the value of tan 15 degree​

Answer 1

Answer:

Tan 15° = Tan(45 – 30)°

By the trigonometry formula, we know,

Tan (A – B) = (Tan A – Tan B) /(1 + Tan A Tan B)

Therefore, we can write,

tan(45 – 30)° = tan 45° – tan 30°/1+tan 45° tan 30°

Now putting the values of tan 45° and tan 30° from the table we get;

tan(45 – 30)° = (1 – 1/√3)/ (1 + 1.1/√3)

tan (15°) = √3 – 1/ √3 + 1

Hence, the value of tan (15°) is √3 – 1/√3 + 1.

We can further resolve the above-resulted expression by putting the value of √3, which is equal to 1.732.

∴ Tan (15°) = (1.732 – 1)/(1.732 + 1) = 0.2679

Or tan (15°) ≈ 0.27

Tan 15° With Respect To sin and cos function

Similarly, we can also find the value of tangent 15 degrees, by knowing the value of sin 15 and cos 15 degrees.

Tan (15°) = sin 15/cos 15

Tan 15° = sin 15/cos 15

Sin 15° = sin (45 – 30)° and cos 15 = cos (45 – 30)°

∴ tan (15°) = sin (45 – 30)° /cos (45 – 30)°

From the trigonometry formulas, we know,

sin(A – B) = sin A cos B – cos A sin B

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

Therefore,

tan (15°)= (sin 45° cos 30° – cos 45° sin 30°)/ (cos 45° cos 30° + sin 45° sin 30°)

Putting the values of sin 30°, sin 45°, cos 30° and cos 45°, we get,

tan 15° = [(1/√2).(√3/2) – (1/√2).(½)] / [(1/√2).(√3/2) + (1/√2).(½)]

Solving the above equation we have,

tan 15° = √3 – 1/ √3 + 1

Step-by-step explanation:

please mark it as brainliest

Answer 2

Answer

$\sf tan 15\degree = tan( 45\degree - 30 \degree)$

we know that,

$\sf tan(A - B) = \dfrac{tanA - tanB}{1 + tanA \: tanB}$

$\sf = \dfrac{ tan45\degree -tan \: 30 \degree}{1 + tan45\degree - tan30 \degree }$

$\sf = \dfrac{ 1 - \dfrac{1}{ \sqrt{3} } }{1 + \dfrac{1}{ \sqrt{3} } }$

$\sf = \dfrac{ \sqrt{3} - 1}{ \sqrt{3} + 1 }$

Now rationalise the denominator,

$\sf = \dfrac{ \sqrt{3} - 1}{ \sqrt{3} + 1 } \times \dfrac{ \sqrt{3} - 1 }{ \sqrt{3} - 1}$

$\sf = \dfrac{ \big(\sqrt{3} - 1 { \big)}^{2} }{ 3 + 1 }$

$\sf = \dfrac{ \big(\sqrt{3} { \big)}^{2} + 1 - 2 \sqrt{3} }{ 2}$

$\sf = \dfrac{ 4 - 2 \sqrt{3} }{ 2}$

$\sf = 2 - \sqrt{3}$

Thus,

$\boxed{ \sf tan \: 15 \degree= 2 - \sqrt{3} }$

Know more :

$\dashrightarrow\sf tan(A + B) = \dfrac{tanA + tanB}{1 - tanA \: tanB}$

$\dashrightarrow\sf tan2A = \dfrac{2tanA }{1 - tan^2A }$

$\dashrightarrow\sf tan3A = \dfrac{3tanA - tan^3A}{1 - 3tan^2A}$

$\bullet\:\sf Trigonometric\:Values :\\\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0& \dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D{e}fined\end{tabular}}$