Question

prove that cos 19 - sin 19 / cos 19 + sin 19 =cot 74

Answer 1

SOLUTION

TO PROVE

$\displaystyle \sf{ \frac{ \cos {19}^{ \circ} - \sin{19}^{ \circ} }{\cos {19}^{ \circ} + \sin{19}^{ \circ} } = \cot {64}^{ \circ} }$

FORMULA TO BE IMPLEMENTED

We are aware of the Trigonometric formula that

$\displaystyle \sf{ \frac{ \cot \alpha \cot \beta - 1}{ \cot \beta + \cot \alpha } = \cot ( \beta + \alpha )}$

PROOF

$\displaystyle \sf{ \frac{ \cos {19}^{ \circ} - \sin{19}^{ \circ} }{\cos {19}^{ \circ} + \sin{19}^{ \circ} } }$

$\displaystyle \sf{ = \frac{ \displaystyle \sf{ \frac{\cos {19}^{ \circ}}{ \sin{19}^{ \circ} } } - \displaystyle \sf{ \frac{\sin {19}^{ \circ}}{ \sin{19}^{ \circ} }}}{ \displaystyle \sf{ \frac{\cos {19}^{ \circ}}{ \sin{19}^{ \circ} } } + \displaystyle \sf{ \frac{\sin {19}^{ \circ}}{ \sin{19}^{ \circ} }} } }$

$\displaystyle \sf{ = \frac{ \cot {19}^{ \circ} - 1}{\cot {19}^{ \circ} + 1 } }$

$\displaystyle \sf{ = \frac{ \cot {19}^{ \circ} \cot {45}^{ \circ} - 1}{\cot {19}^{ \circ} + \cot {45}^{ \circ} } }$

$\displaystyle \sf{ = \cot ({19}^{ \circ} + {45}^{ \circ}) }$

$\displaystyle \sf{ = \cot {64}^{ \circ} }$

Hence proved

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Answer 2

Answer:

SOLUTION

TO PROVE

\displaystyle \sf{ \frac{ \cos {19}^{ \circ} - \sin{19}^{ \circ} }{\cos {19}^{ \circ} + \sin{19}^{ \circ} } = \cot {64}^{ \circ} }

cos19

∘

+sin19

∘

cos19

∘

−sin19

∘

=cot64

∘

FORMULA TO BE IMPLEMENTED

We are aware of the Trigonometric formula that

\displaystyle \sf{ \frac{ \cot \alpha \cot \beta - 1}{ \cot \beta + \cot \alpha } = \cot ( \beta + \alpha )}

cotβ+cotα

cotαcotβ−1

=cot(β+α)

PROOF

\displaystyle \sf{ \frac{ \cos {19}^{ \circ} - \sin{19}^{ \circ} }{\cos {19}^{ \circ} + \sin{19}^{ \circ} } }

cos19

∘

+sin19

∘

cos19

∘

−sin19

∘

\displaystyle \sf{ = \frac{ \displaystyle \sf{ \frac{\cos {19}^{ \circ}}{ \sin{19}^{ \circ} } } - \displaystyle \sf{ \frac{\sin {19}^{ \circ}}{ \sin{19}^{ \circ} }}}{ \displaystyle \sf{ \frac{\cos {19}^{ \circ}}{ \sin{19}^{ \circ} } } + \displaystyle \sf{ \frac{\sin {19}^{ \circ}}{ \sin{19}^{ \circ} }} } }=

sin19

∘

cos19

∘

+

sin19

∘

sin19

∘

sin19

∘

cos19

∘

−

sin19

∘

sin19

∘

\displaystyle \sf{ = \frac{ \cot {19}^{ \circ} - 1}{\cot {19}^{ \circ} + 1 } }=

cot19

∘

+1

cot19

∘

−1

\displaystyle \sf{ = \frac{ \cot {19}^{ \circ} \cot {45}^{ \circ} - 1}{\cot {19}^{ \circ} + \cot {45}^{ \circ} } }=

cot19

∘

+cot45

∘

cot19

∘

cot45

∘

−1

\displaystyle \sf{ = \cot ({19}^{ \circ} + {45}^{ \circ}) }=cot(19

∘

+45

∘

)

\displaystyle \sf{ = \cot {64}^{ \circ} }=cot64

∘

Hence proved