Question-216230

Question Number 216230 by MATHEMATICSAM last updated on 31/Jan/25

Answered by mahdipoor last updated on 31/Jan/25

\frac{{s e c}^{2} - t a n}{{s e c}^{2} + t a n} \times \frac{{c o s}^{2}}{{c o s}^{2}} = \frac{1 - s i n . c o s}{1 + s i n . c o s} = \frac{2 - s i n (2 x)}{2 + s i n (2 x)}

= \frac{A}{B} 1 < A, B < 3 \Rightarrow m i n, m a x = \frac{1}{3}, \frac{3}{1}