# Prove that Cosec (67° + θ) − Sec (23° − θ) = 0 - Mathematics

Sum

Prove that

cosec (67° + θ) − sec (23° − θ) = 0

#### Solution

$\begin{array}{l} \text{L.H.S}=cosec( {67}^\circ + \theta) - \sec( {23}^\circ- \theta) \\ \end{array}$

$\begin{array}{l}=cosec{ {90}^\circ - ( {23}^\circ-\theta)}-\sec( {23}^\circ - \theta) \\ \end{array}$

$\begin{array}{l}=\sec( {23}^\circ -\theta) - \sec( {23}^\circ - \theta) \\ \\ \end{array}$

$=0=R.H.S.$

Concept: Trigonometric Ratios of Some Special Angles
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#### APPEARS IN

RS Aggarwal Secondary School Class 10 Maths
Chapter 7 Trigonometric Ratios of Complementary Angles
Exercises | Q 7.3 | Page 314

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