Question

(i) $sec\theta (1-sin\theta )(sec\theta +tan\theta )=1$
(ii) $sin\theta (1+tan\theta )+cos\theta (1+cot\theta )=(sec\theta +cosec\theta )$

Answer 1

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

(ii) $sin\theta (1+tan\theta )+cos\theta (1+cot\theta )=(sec\theta +cosec\theta )LHS=sin\theta (1+tan\theta )+cos\theta (1+cot\theta )=sin\theta +sin\theta tan\theta +cos\theta +cos\theta cot\theta =sin\theta +sin\theta \times \frac{sin\theta }{cos\theta }+cos\theta +cos\theta \times \frac{cos\theta }{sin\theta }=sin\theta +cos\theta +\frac{si{n}^2\theta }{cos\theta }+\frac{co{s}^2\theta }{sin\theta }=(sin\theta +cos\theta )+\frac{si{n}^3\theta +co{s}^3\theta }{sin\theta cos\theta }=\frac{(sin\theta +cos\theta )sin\theta cos\theta +(si{n}^3\theta +co{s}^3\theta )}{sin\theta cos\theta }=\frac{si{n}^2\theta cos\theta +co{s}^2\theta sin\theta +si{n}^3\theta +co{s}^3\theta }{sin\theta cos\theta }=\frac{(si{n}^2\theta cos\theta +co{s}^3\theta )+(co{s}^2\theta sin\theta +si{n}^3\theta )}{sin\theta cos\theta }=\frac{cos\theta (si{n}^2\theta +co{s}^2\theta )+sin\theta (si{n}^2\theta +co{s}^2\theta )}{sin\theta cos\theta }=\frac{cos\theta +sin\theta }{sin\theta cos\theta }=\frac{cos\theta }{sin\theta cos\theta }+\frac{sin\theta }{sin\theta cos\theta }=\frac{1}{sin\theta }+\frac{1}{cos\theta }=(sec\theta +cosec\theta )=RHS$