Question

If sin40°=3x-2, evaluate tan50°+tan20°=? in terms of x

Answer 1

SOLUTION

GIVEN

sin 40° = 3x - 2

TO DETERMINE

tan 50° + tan 20° in terms of x

EVALUATION

Here it is given that sin 40° = 3x - 2

Now

$\displaystyle \sf{ \tan {50}^{ \circ} + \tan {20}^{ \circ}}$

$\displaystyle \sf{ = \frac{\sin {50}^{ \circ}}{\cos {50}^{ \circ}} + \frac{\sin {20}^{ \circ}}{\cos {20}^{ \circ}} }$

$\displaystyle \sf{ = \frac{\sin {50}^{ \circ}\cos {20}^{ \circ} +\cos {50}^{ \circ}\sin {20}^{ \circ} }{\cos {50}^{ \circ}\cos {20}^{ \circ}} }$

$\displaystyle \sf{ = \frac{\sin ({50}^{ \circ} + {20}^{ \circ} )}{\cos {50}^{ \circ}\cos {20}^{ \circ}} }$

$\displaystyle \sf{ = \frac{\sin {70}^{ \circ} }{\cos {50}^{ \circ}\cos {20}^{ \circ}} }$

$\displaystyle \sf{ = \frac{\sin ({90}^{ \circ} - {20}^{ \circ} )}{\cos {50}^{ \circ}\cos {20}^{ \circ}} }$

$\displaystyle \sf{ = \frac{\cos {20}^{ \circ} }{\cos {50}^{ \circ}\cos {20}^{ \circ}} }$

$\displaystyle \sf{ = \frac{1}{\cos {50}^{ \circ}} }$

$\displaystyle \sf{ = \frac{1}{\cos ({90}^{ \circ} - {40}^{ \circ})} }$

$\displaystyle \sf{ = \frac{1}{\sin {40}^{ \circ} } }$

$\displaystyle \sf{ = \frac{1}{3x - 2} }$

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