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Prove the following: tan θ + tan (90° – θ) = sec θ sec (90° – θ)

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Solution. 
tan θ + tan (90° – θ) = sec θ sec (90° – θ)
  Taking L.H.S.
= tan θ + tan (90° – θ)
= tanθ + cotθ    (\because tan (90 – θ) = cot θ)
  =\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }      \left ( \because \tan \theta = \frac{\sin \theta }{\cos \theta },\cot \theta = \frac{\cos \theta }{\sin \theta } \right )
Taking L.C.M.
\frac{\sin ^{2}\theta +\cos ^{2}\theta }{\sin \theta \cos \theta }
= \frac{1}{\cos \theta \cdot \sin \theta }       \left ( \because \sin ^{2}\theta +\cos ^{2}\theta = 1 \right )
= \frac{1}{\cos \theta }\times \frac{1}{\sin \theta }
= \sec \theta \times \cos ec\, \theta     \left ( \because \frac{1}{\cos \theta } = \sec \theta ,\frac{1}{\sin \theta }= \cos ec\, \theta \right )
= \sec \theta \times \sec \left ( 90^{\circ}-\theta \right )          \left ( \because \sec \left ( 90-\theta \right ) = \cos ec\, \theta \right )
L.H.S. = R.H.S.
Hence proved.
 

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