Question

show that tan 75+ cot 75=4​

Answer 1

Answer: HOPE THIS HELPS YOU

Step-by-step explanation: Show that tan 75° + cot 75° = 4.

Solution :

tan 75°  =  tan (45° + 30°)

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

 =  (1 + (1/√3)) /(1 - 1(1/√3))

 =  [(√3 + 1)/√3] / [(√3 - 1)/√3]

 =  (√3 + 1)/(√3 - 1)

Multiply by (√3 + 1) on both numerator and denominator.

 =  (√3 + 1)2/(√32 - 12)

 =  (3 + 1 + 2√3) / (3 - 1)

 =  (4 + 2√3) / 2

 =  2 + √3   ------(1)

tan 15°  =  cot (90° - 15°)

tan 15°  =  cot 75°

Instead of finding the value of cot 75, let us find the value of tan 15.

tan 15°  =  tan (45° - 30°)

Using compound angle formula, we get

 =  (√3 - 1)/(√3 + 1)

Multiply by (√3 - 1) on both numerator and denominator.

 =  (√3 - 1)2/(√32 - 12)

 =  (3 + 1 - 2√3) / (3 - 1)

 =  (4 - 2√3) / 2

 =  2 - √3   ------(2)

(1) + (2)

tan 75° + cot 75°  =    2 + √3  +  2 - √3  

tan 75° + cot 75°  =  4

Hence proved.

Answer 2

$\rm \huge { \underline{ \underline{ \blue { answer : }}}}$

$\bf \huge\fbox \green{ \: \: tan \:( A + B) = \frac{tan \:A + tan B }{1 -tan A \times tan B} \: }$

$\bf \: cot \: A = \frac{1}{tanA}$

$\bf \: {tan \: 75° = tan(30° + 45°)} \\ \bf \: = \frac{ \tan(30° + \tan(45°) ) }{1 - \tan(30° \times \tan(45°) ) } \\ \bf \: = \frac{ \frac{1}{ \sqrt{3} + 1 } }{ \frac{1 - 1}{ \sqrt{3} \times 1 } } \\ \bf \: = \frac{ \sqrt{3} + 1 }{ \sqrt{3} - 1 }$

$\bf \: and \\ \rm \huge \fbox \red{ \cot(75°) = \frac{1}{ \tan(75°) } }$

$\bf \: \implies \: cot75° = \frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1 }$

$\bf \: \tan(75°) + \cot(75°) = \frac{ \sqrt{3} + 1}{ \sqrt{3} - 1 } + \frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1} \\ \bf \: = \frac{ {( \sqrt{3} + 1 )}^{2} + {( \sqrt{3} - 1)}^{2} }{( \sqrt{3} - 1)( \sqrt{3} + 1)} \\ \bf \: \frac{(3 + 1 + 2 \sqrt{3}) + (3 + 1 - 2 \sqrt{3} ) }{3 - 1}$

$\bf \: = \frac{3 + 1 + 3 + 1}{2} \\ \bf \: = \frac{8}{2} \\ \bf \: = 4$

$\rm \huge \pink{hence \: proved}$

@sanju