Question

tan^2 a + sec a = 5 ,find cos a​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$\huge\bold \red{\boxed{ANSWER}}$

$\star \: \: \: \blue{solution : }$

• ${tan}^{2} \theta + sec \theta = 5 \\ = > {sec}^{2} \theta - 1 + sec \theta = 5\\ = > {sec}^{2} \theta + sec \theta - 6 = 0 \\ = > {sec}^{2} \theta + 3 \: sec \theta - 2sec \theta - 6 = 0 \\ = > sec \theta(sec \theta + 3) - 2(sec \theta + 3) = 0 \\ = > (sec \theta + 3)(sec \theta - 2) = 0 \\ \\ sec \theta = - 3 \\ sec \theta = 2 \: \: \: \\ \\ {\boxed{\sec{\theta = 2 \: \: \: \: = > \theta = 60}}}$

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