Question

cos 27 + sin 12 .(a) Show that cos 12-sin12 = Tan 57°​

Answer 1

$\displaystyle \sf{ \frac{cos \: {12}^{ \circ} +sin \: {12}^{ \circ} }{cos \: {12}^{ \circ} - sin \: {12}^{ \circ}} = tan {57}^{ \circ} }$

Given :

$\displaystyle \sf{ \frac{cos \: {12}^{ \circ} +sin \: {12}^{ \circ} }{cos \: {12}^{ \circ} - sin \: {12}^{ \circ}} = tan {57}^{ \circ} }$

To find : To prove

Solution :

We are aware of the Trigonometric formula that

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

Now we have

RHS

$\displaystyle \sf{ = tan {57}^{ \circ} }$

$\displaystyle \sf{ = tan ( {45}^{ \circ} + {12}^{ \circ}) }$

$\displaystyle \sf{ = \frac{tan \: {45}^{ \circ} +tan \: {12}^{ \circ}}{1 -tan \: {45}^{ \circ} tan \: {12}^{ \circ}} }$

$\displaystyle \sf{ = \frac{1 +tan \: {12}^{ \circ}}{1 - tan \: {12}^{ \circ}} }$

$\displaystyle \sf{ = \frac{1 + \dfrac{sin \: {12}^{ \circ}}{cos \: {12}^{ \circ}} }{1 - \dfrac{sin \: {12}^{ \circ}}{cos \: {12}^{ \circ}}} }$

$\displaystyle \sf{ = \frac{ \dfrac{cos \: {12}^{ \circ} + sin \: {12}^{ \circ}}{cos \: {12}^{ \circ}} }{ \dfrac{cos \: {12}^{ \circ} - sin \: {12}^{ \circ}}{cos \: {12}^{ \circ}}} }$

$\displaystyle \sf{ = \frac{ cos \: {12}^{ \circ} + sin \: {12}^{ \circ}}{ cos \: {12}^{ \circ} - sin \: {12}^{ \circ}} }$

= LHS

Hence the proof follows

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