First slide

Trigonometric Identities

Question

If $a \sec θ + b \tan θ = c$ then ${(a \tan θ + b \sec θ)}^{2} =$

Easy

A

$a^{2} + b^{2} - c^{2}$
B

$- a^{2} + b^{2} + c^{2}$
C

$a^{2} + b^{2} + c^{2}$
D

$a^{2} - b^{2} + c^{2}$

Solution

$G i v e n a s e c θ + b \tan θ = c .$
$\begin{array}{l} L e t a \tan θ + b \sec θ = k \\ S q u a r i n g a n d s u b t r a c t i n g o n b o t h s i d e s, w e g e t \\ ∴ a^{2} - b^{2} = c^{2} - k^{2} (∵ s e c^{2} θ - \tan^{2} θ = 1) \\ \Rightarrow k^{2} = c^{2} - a^{2} + b^{2} \\ = - a^{2} + b^{2} + c^{2} \end{array}$

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