Question

Express cosec 69°+cot 69°n terms of trigonometric ratios of angles between 0° and 45°​

Answer 1

Question :-

Express cosec 69°+cot 69°n terms of trigonometric ratios of angles between 0° and 45°

Solution :-

We know that,

• cos A = sin(90° - A)
• cot A = tan(90° - A)
• cosec A = sec(90° - A)

By using,

$\sf:\longmapsto{ \cosec 69 \degree = \sec(90 \degree - 69 \degree) } \\ \\ \bf:\longmapsto \red{\cosec 69 \degree = \sec21 \degree } \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf:\longmapsto{ \cot69 \degree = \tan(90 \degree - 69 \degree) } \: \: \: \\ \\ \bf:\longmapsto \red{ \cos69 \degree = \tan21 \degree } \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Hence,

$\bf:\longmapsto \red{ \cosec69 \degree + \cot69 \degree = \sec21 \degree + \tan21 \degree }$

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Trigonometry Table :-

$\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}$

Answer 2

We know that,

• cos A = sin(90° - A)
• cot A = tan(90° - A)
• cosec A = sec(90° - A)

By using,

⟼cosec69°=sec(90°−69°)

⟼cosec69°=sec21°

:⟼cot69°=tan(90°−69°)

:⟼cos69°=tan21°

Hence

⟼cosec69°+cot69°=sec21°+tan21°

Trigonometry Table :-

Refer to the attachment