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Prove that Tan (55° − θ) − Cot (35° + θ) = 0 - Mathematics

Sum

Prove that

tan (55° − θ) − cot (35° + θ) = 0

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Solution

\[\begin{array}{l} LHS =\tan( {55}^0 - \theta) - \cot( {35}^0 + \theta) \\ \end{array}\]

\[\begin{array}{l}=\tan{ {90}^0  - ( {35}^0 + \theta)} - \cot( {35}^0 + \theta) \\ \end{array}\]

\[\begin{array}{l}=\cot( {35}^0 + \theta) - \cot( {35}^0 + \theta) \\ \end{array}\]

= 0

= RHS

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APPEARS IN

RS Aggarwal Secondary School Class 10 Maths
Chapter 7 Trigonometric Ratios of Complementary Angles
Exercise | Q 7.2 | Page 314
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