InterviewSolution
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What is the value of \(\frac{{\cot x}}{{1 - \tan x}} + \frac{{\tan x}}{{1 - \cot x}}?\)1). sin x cos x + 12). sec x cosec x + 13). tan x cot x + 14). sec2x cosec2 x + 1 |
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Answer» $({{\COT x}}{{1 - \tan x}} + \frac{{\tan x}}{{1 - \cot x}} = \frac{{cosx}}{{\frac{{sinx}}{{1 - \frac{{sinx}}{{cosx}}}}}} + \frac{{sinx}}{{\frac{{cosx}}{{1 - \frac{{cosx}}{{sinx}}}}}} = \frac{{cosx}}{{\frac{{sinx}}{{\frac{{cosx - sinx}}{{cosx}}}}}} + \frac{{sinx}}{{\frac{{cosx}}{{\frac{{sinx - cosx}}{{sinx}}}}}})$ $(= \frac{{co{s^2}x}}{{sinx\left( {cosx - sinx} \RIGHT)}} + \frac{{si{n^2}x}}{{cosx\left( {sinx - cosx} \right)}})$ $(= \frac{{co{s^2}x}}{{sinx\left( {cosx - sinx} \right)}} - \frac{{si{n^2}x}}{{cosx\left( {cosx - sinx} \right)}})$ $(= \frac{{co{s^3}x - si{n^3}x}}{{\left( {cosx - sinx} \right)sinx\;cosx}} = \frac{{\left[ {\left( {cosx - sinx} \right)\left( {si{n^2}x +co{s^2}x\; + \;sinx\;cosx} \right)} \right]}}{{\left( {cosx - sinx} \right)\sin cosx}}\;)$ $(= \frac{{1 +sinx\;cosx}}{{SINXCOSX}})$ $(= COSECX\;secx + 1)$ |
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