Question

prove that (tan²y) (cos²y) + 1/sec²y = 1​

Answer 1

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identities

tan x = $\frac{sinx}{cosx}$ , sec x = $\frac{1}{cosx}$ , sin²x + cos²x = 1

Consider the left side

tan²y × cos²y + $\frac{1}{sec^2y}$

= $\frac{sin^2y}{cos^2y}$ × cos²y + $\frac{1}{\frac{1}{cos^2y} }$ ← cancel cos²y in the product

= sin²y + cos²y

= 1

= right side ⇒ proven

Answer 2

Step-by-step explanation:

$(\tan^2y) (\cos^2y) +\frac{1}{\sec^2y} = 1\\

LHS

= (\tan^2y) (\cos^2y) +\frac{1}{\sec^2y} \\\\

\frac{\sin^2y}{\cos^2y}\times \cos^2 y + \cos^2 y \\\\

= \sin^2y + \cos^2 y \\\\

= 1\\\\

= RHS\\\\

\purple {\boxed {\bold {\therefore (\tan^2y) (\cos^2y) +\frac{1}{\sec^2y} = 1}}}$

Hence Proved