Question

If tan show that tan 160^ -tan 110^ 1+tan 160^ tan 110^ = 1-k^ 2 2k .​

Answer 1

Given:

If tan 20° = K, then show that $\sf\dfrac {tan \: 160° \: - \: tan \: 110°}{1 \: + \: tan \: 160° \: . \: tan \: 110°} \: = \: \dfrac {1 \: - \: {k}^{2}}{2k}$

Solution:

$\sf\ \: = \: \: \: \dfrac {tan \: 160° \: - \: tan \: 110°}{1 \: + \: tan \: 160° \: . \: tan \: 110°} \: = \: \dfrac {1 \: - \: {k}^{2}}{2k}$

$\sf\ \: = \: \: \: \dfrac {tan \: ( \: 180 \: - \: 20 \: ) \: - \: tan \: ( \: 90 \: + \: 20 \: )}{1 \: + \: tan \: ( \: 180 \: - \: 20 \: ) \: . \: tan \: ( \: 90 \: + \: 20 \: )}$

$\sf\ \: = \: \: \: \dfrac { \: - \: tan \: 20° \: - \: ( \: - \: cot \: 20° \: )}{1 \: + \: ( \: - \: tan \: 20° \: ) \: \: ( \: - \: cot \: 20°)}$

$\sf\ \: = \: \: \: \dfrac { \: - \: tan \: 20° \: + \: cot \: 20°}{1 \: + \: tan \: 20° \: cot \: 20°}$

$\sf\ \: = \: \: \: \dfrac{ - \: k \: + \: \frac{1}{k}}{ \: \: 1 \: + \: 1}$

$\sf\ \: = \: \: \: \dfrac{ \frac{ - \: {k}^{2} \: + \: 1}{k}}{ \frac{2}{1}}$

$\sf\ \: = \: \: \: \dfrac{ {1 \: - \: k}^{2}}{k} \: \times \: \dfrac{1}{2}$

$\: \:{\red{ \: = }} \: \: \: \: {\sf{\red{\dfrac{{1 \: - \: k}^{2}}{2k}}}}$

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Additional Information:

Values for Sin :

• sin (90° - $\theta$) = + cos $\theta$
• sin (90° + $\theta$) = + cos $\theta$
• sin (180° - $\theta$) = + sin $\theta$
• sin (180° + $\theta$) = - sin $\theta$
• sin (270° - $\theta$) = - cos $\theta$
• sin (270° + $\theta$) = - cos $\theta$
• sin (360° - $\theta$) = - sin $\theta$
• sin (360° + $\theta$) = + sin $\theta$

Values for Cos :

• cos (90° - $\theta$) = + sin $\theta$
• cos (90° + $\theta$) = - sin $\theta$
• cos (180° - $\theta$) = - cos $\theta$
• cos (180° + $\theta$) = - cos $\theta$
• cos (270° - $\theta$) = - sin $\theta$
• cos (270° + $\theta$) = + sin $\theta$
• cos (360° - $\theta$) = + cos $\theta$
• cos (360° + $\theta$) = + cos $\theta$

Values for tan :

• tan (90° - $\theta$) = + cot $\theta$
• tan (90° + $\theta$) = - sin $\theta$
• tan (180° - $\theta$) = - tan $\theta$
• tan (180° + $\theta$) = + tan $\theta$
• tan (270° - $\theta$) = + cot $\theta$
• tan (270° + $\theta$) = - cot $\theta$
• tan (360° - $\theta$) = - tan $\theta$
• tan (360° + $\theta$) = + tan $\theta$

Values for cot :

• cot (90° - $\theta$) = + tan $\theta$
• cot (90° + $\theta$) = - tan $\theta$
• cot (180° - $\theta$) = - cot $\theta$
• cot (180° + $\theta$) = + cot $\theta$
• cot (270° - $\theta$) = + tan $\theta$
• cot (270° + $\theta$) = - tan $\theta$
• cot (360° - $\theta$) = - cot $\theta$
• cot (360° + $\theta$) = + cot $\theta$

Values for csc :

• csc (90° - $\theta$) = + sec $\theta$
• csc (90° + $\theta$) = + sec $\theta$
• csc (180° - $\theta$) = + csc $\theta$
• csc (180° + $\theta$) = - csc $\theta$
• csc (270° - $\theta$) = - sec $\theta$
• csc (270° + $\theta$) = - sec $\theta$
• csc (360° - $\theta$) = - csc $\theta$
• csc (360° + $\theta$) = + csc $\theta$

Values for sec :

• sec (90° - $\theta$) = + csc $\theta$
• sec (90° + $\theta$) = - csc $\theta$
• sec (180° - $\theta$) = - sec $\theta$
• sec (180° + $\theta$) = - sec $\theta$
• sec (270° - $\theta$) = - csc $\theta$
• sec (270° + $\theta$) = + csc $\theta$
• sec (360° - $\theta$) = - sec $\theta$
• sec (360° + $\theta$) = + cot $\theta$

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

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