Question

(i) $\frac{sec\theta -1}{sec\theta +1}=\frac{si{n}^2\theta }{(1+cos\theta {)}^2}$

(ii) $\frac{sec\theta -tan\theta }{sec\theta +tan\theta }=\frac{co{s}^2\theta }{(1+sin\theta {)}^2}$

Answer 1

(i) $\frac{sec\theta -1}{sec\theta +1}=\frac{si{n}^2\theta }{(1+cos\theta {)}^2}L.H.S=\frac{sec\theta -1}{sec\theta +1}=\frac{(\frac{1}{cos\theta }-1)}{(\frac{1}{cos\theta }+1)}=\frac{\frac{1–cos\theta }{cos\theta }}{\frac{1+cos\theta }{cos\theta }}=\frac{1–cos\theta }{1+cos\theta }Multiplyingnumeratoranddenominatorby(1+cos\theta ),weget=\frac{1-cos\theta \left)\right(1+cos\theta )}{(1+cos\theta {)}^2}=\frac{(1–co{s}^2\theta )}{(1+cos\theta {)}^2}=\frac{si{n}^2\theta }{(1+cos\theta {)}^2}=R.H.S$

Hence proved

(ii) $\frac{sec\theta -tan\theta }{sec\theta +tan\theta }=\frac{co{s}^2\theta }{(1+sin\theta {)}^2}LHS=\frac{sec\theta -tan\theta }{sec\theta +tan\theta }\text{Rationalizing the denominator by multiplying and dividing with}sec\theta +tan\theta \text{, we get}LHS=\frac{sec\theta -tan\theta }{sec\theta +tan\theta }\times \frac{sec\theta +tan\theta }{sec\theta +tan\theta }=\frac{se{c}^2\theta -ta{n}^2\theta }{(sec\theta +tan\theta {)}^2}=\frac{1}{se{c}^2\theta +ta{n}^2\theta +2sec\theta tan\theta }=\frac{1}{\frac{1}{co{s}^2\theta }+\frac{si{n}^2\theta }{co{s}^2\theta }+2\times \frac{1}{cos\theta }\times \frac{sin\theta }{cos\theta }}=\frac{1}{\frac{1+si{n}^2\theta +2sin\theta }{co{s}^2\theta }}=\frac{co{s}^2\theta }{1+si{n}^2\theta +2sin\theta }=\frac{co{s}^2\theta }{(1+sin\theta {)}^2}=RHS$

Hence proved