Question

tan 15°+tan 75°+cot 15°+cot 75°
options
1)4√3
2)√2
3)√6
4)8​

Answer 1

Answer:

$\boxed{\boxed{\huge { \pink{8}}}}$

Step-by-step explanation:

$tan75 \degree = tan(90 \degree - 15 \degree) \\tan75 \degree = cot15 \degree \\ cot75 \degree = cot(90 \degree - 15 \degree) \\ cot75 \degree = tan15 \degree \\ tan15 \degree = tan(45 \degree - 30 \degree) \\ \\ = \dfrac{tan45 \degree - tan30 \degree}{1 + tan45 \degree \: tan30 \degree} \\ \\ = \dfrac{1 - \dfrac{1}{ \sqrt{3} } }{1 + 1 \: \times \dfrac{1}{ \sqrt{3} } } \\ \\ = \dfrac{ \dfrac{ \sqrt{3} - 1 }{ \sqrt{3} } }{ \dfrac{ \sqrt{3} + 1}{ \sqrt{3} } } \\ \\ = \frac{ \sqrt{3} - 1}{ \sqrt{3} + 1} \\ = \dfrac{( \sqrt{3} - 1) \: ( \sqrt{3} - 1) }{( \sqrt{3} + 1) \: ( \sqrt{3} - 1)} \\ = \dfrac{ {( \sqrt{3} - 1)}^{2} }{ {( \sqrt{3} )}^{2} - {(1)}^{2} } \\ = \dfrac{ {( \sqrt{3} \: )}^{2} + {(1)}^{2} - 2(1) \: ( \sqrt{3} )}{3 - 1} \\ = \dfrac{3 + 1 - 2 \sqrt{3} }{2} \\ = \dfrac{4 - 2 \sqrt{3} }{2} \\ = \frac{ 2(2 - \sqrt{3} )}{2} \\ tan15 \degree = 2 - \sqrt{3} \\ cot15 \degree = \dfrac{1}{tan15 \degree} \\ = \dfrac{1}{2 - \sqrt{3} } \\ = \dfrac{(2 + \sqrt{3} )}{(2 - \sqrt{3}) \: (2 + \sqrt{3}) } \\ = \dfrac{2 + \sqrt{3} }{ {(2)}^{2} - { (\sqrt{3}) }^{2} } \\ = \dfrac{2 + \sqrt{3} }{4 - 3} \\ = \dfrac{2 + \sqrt{3} }{1} \\ cot15 \degree = 2 + \sqrt{3}$

$tan15 \degree + tan75 \degree + cot15 \degree + \cot75 \degree \\ = tan15 \degree + cot15 \degree + cot15 \degree + tan15 \degree \\ = 2 \: tan15 \degree + 2 \: cot15 \degree \\ = 2 \: (tan15 \degree + cot15 \degree) \\ = 2 \: (2 - \sqrt{3} + 2 + \sqrt{3} ) \\ = 2 \: (4) \\ = 8$

Answer 2

$\large\underline{\sf{Solution-}}$

We know,

$\red{\rm :\longmapsto\:tan(90 \degree \: - \theta ) = cot\theta}$

and

$\red{\rm :\longmapsto\:cot(90 \degree \: - \theta ) = tan\theta}$

Now,

Consider,

$\rm :\longmapsto\:tan15\degree \: + tan75\degree \: + cot15\degree \: + cot75\degree \:$

can be rewritten as

$\rm \: = \: tan15\degree + cot15\degree + tan(90\degree -15\degree) + cot(90 - 15\degree)$

$\rm \: = \: tan15\degree \: + cot15\degree \: + cot15\degree \: + tan15\degree \:$

$\rm \: = \: \: 2(tan15\degree \: + cot15\degree \: )$

$\rm \: = \: \: 2\bigg(\dfrac{sin15\degree \: }{cos15\degree \: } + \dfrac{cos15\degree \: }{sin15\degree \: } \bigg)$

$\rm \: = \: \: 2\bigg(\dfrac{sin^{2} 15\degree + {cos}^{2} 15\degree \: }{cos15\degree \: sin15\degree \: } \bigg)$

We know,

$\blue{ \boxed{\bf \: {sin}^{2}x + {cos}^{2}x = 1}}$

So, using this

$\rm \: = \: \: \dfrac{2 \times 1}{sin15\degree \: cos15\degree \: }$

$\rm \: = \: \: \dfrac{2}{sin15\degree \: cos15\degree \: }$

On multiply numerator and denominator by 2, we get

$\rm \: = \: \: \dfrac{4}{2 \: sin15\degree \: cos15\degree \: }$

We know,

$\blue{ \boxed{\bf \:sin2x = 2 \: sinx \: cosx}}$

So, using this we have

$\rm \: = \: \: \dfrac{4}{sin( \: 2 \times 15\degree)}$

$\rm \: = \: \: \dfrac{4}{sin30\degree \: }$

$\rm \: = \: \: 4 \times 2$

$\rm \: = \: \: 8$

Hence,

$\bf :\longmapsto\:tan15\degree + tan75\degree + cot15\degree + cot75\degree = 8$

Hence,

Option (4) is correct.

Additional Information :-

Trigonometry Formulas

sin(−θ) = −sin θ

cos(−θ) = cos θ

tan(−θ) = −tan θ

cosec(−θ) = −cosecθ

sec(−θ) = sec θ

cot(−θ) = −cot θ

Product to Sum Formulas

sin x sin y = 1/2 [cos(x–y) − cos(x+y)]

cos x cos y = 1/2[cos(x–y) + cos(x+y)]

sin x cos y = 1/2[sin(x+y) + sin(x−y)]

cos x sin y = 1/2[sin(x+y) – sin(x−y)]

Sum to Product Formulas

sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]

sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]

cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]

cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]

Sum or Difference of angles

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

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

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

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

tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)]

tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)]

cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)]

cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)]

cos(A+B) cos(A–B)=cos^2A–sin^2B=cos^2B–sin^2A

sin(A+B) sin(A–B) = sin^2A–sin^2B=cos^2B–cos^2A

Multiple and Submultiple angles

sin2A = 2sinA cosA = [2tan A /(1+tan²A)]

cos2A = cos²A–sin²A = 1–2sin²A = 2cos²A–1= [(1-tan²A)/(1+tan²A)]

tan 2A = (2 tan A)/(1-tan²A)