Question

find the solution of tan 15/2

Answer 1

value of tan(15/2)° is √6 - √3 + √2 - 2

we have to find tan(15/2)°

using formula, tanθ = √{(1 - cos2θ)/(1 + cos2θ)}

Let θ = 15/2° ⇒2θ = 15°

then, tan(15/2)° = √{(1 - cos15°)/(1 + cos15°)}

we know cos15° = cos(45 - 30)

= cos45°.cos30° + sin45°.sin30°

= √3/2√2√2 + 1/2√2

= (√3 + 1)/2√2

tan(15/2)° = √[{1 - (√3 + 1)/2√2}/{1 + (√3 + 1)/2√2}]

= √{(2√2 - √3 - 1)/(2√2 + √3 + 1)}

after rationalisation we get,

= √{(2√2)² - (√3 + 1)²/(2√2 + √3 + 1)²}

= √{(8 - 4 - 2√3)/(2√2 + √3 + 1)²}

= √{(√3 - 1)²/(2√2 + √3 + 1)²}

= (√3 - 1)/(2√2 + √3 + 1)

again rationalisation we get,

= √6 - √3 + √2 - 2

Answer 2

Answer:

Step-by-step explanation:

value of tan(15/2)° is √6 - √3 + √2 - 2

we have to find tan(15/2)°

using formula, tanθ = √{(1 - cos2θ)/(1 + cos2θ)}

Let θ = 15/2° ⇒2θ = 15°

then, tan(15/2)° = √{(1 - cos15°)/(1 + cos15°)}

we know cos15° = cos(45 - 30)

= cos45°.cos30° + sin45°.sin30°

= √3/2√2√2 + 1/2√2

= (√3 + 1)/2√2

tan(15/2)° = √[{1 - (√3 + 1)/2√2}/{1 + (√3 + 1)/2√2}]

= √{(2√2 - √3 - 1)/(2√2 + √3 + 1)}

after rationalisation we get,

= √{(2√2)² - (√3 + 1)²/(2√2 + √3 + 1)²}

= √{(8 - 4 - 2√3)/(2√2 + √3 + 1)²}

= √{(√3 - 1)²/(2√2 + √3 + 1)²}

= (√3 - 1)/(2√2 + √3 + 1)

again rationalisation we get,

= √6 - √3 + √2 - 2