Question

1/sec
theta - tan theta = sec theta + tan theta​

Answer 1

Answer:

dont get that question

I am unable to understand

Answer 2

Answer:

Given:–

→ $\frac{1}{secθ - tanθ} = secθ + tanθ$

Take LHS,

→ $\frac{1}{secθ - tanθ}$

On rationalizing,

→ $\frac{1}{secθ - tanθ} \times \frac{secθ + </strong><strong> </strong><strong> </strong><strong> </strong><strong>tanθ}{secθ + tanθ}$

→ $\frac{secθ + tanθ}{ {sec}^{2}θ - {tan}^{2} θ }$

Since,

sec²θ-tan²θ=1

→ $secθ + tanθ$