Question Number 86231 by liki last updated on 27/Mar/20
Commented by liki last updated on 27/Mar/20
$$..{plz}\:{i}\:{need}\:{help}\:{qn}\:{no}.\:\mathrm{5}\left({a}\right)\:\&\:{c} \\ $$
Answered by TANMAY PANACEA. last updated on 27/Mar/20
$$\frac{{tan}\mathrm{145}−{tan}\mathrm{125}}{\mathrm{1}+{tan}\mathrm{145}{tan}\mathrm{125}} \\ $$$$={tan}\left(\mathrm{145}−\mathrm{125}\right) \\ $$$$={tan}\mathrm{20} \\ $$$$={cot}\mathrm{70} \\ $$$$=\frac{\mathrm{1}}{{tan}\mathrm{70}} \\ $$$$=\frac{\mathrm{1}−{tan}^{\mathrm{2}} \mathrm{35}}{\mathrm{2}{tan}\mathrm{35}} \\ $$$$=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{2}{x}} \\ $$
Commented by liki last updated on 27/Mar/20
$$….{Thank}\:{you}\:{so}\:{much} \\ $$
Answered by TANMAY PANACEA. last updated on 27/Mar/20
$${cosx}+{cosy}={cosa} \\ $$$${sinx}+{siny}={sina} \\ $$$${a}=\frac{\mathrm{7}\pi}{\mathrm{12}}=\mathrm{105}^{{o}} \\ $$$$\left.{from}\:\mathrm{1}\right)\:\:\mathrm{2}{cos}\frac{{x}+{y}}{\mathrm{2}}{cos}\frac{{x}−{y}}{\mathrm{2}}={cosa} \\ $$$$\left.{from}\:\mathrm{2}\right)\:\mathrm{2}{sin}\frac{{x}+{y}}{\mathrm{2}}{cos}\frac{{x}−{y}}{\mathrm{2}}={sina} \\ $$$${tan}\left(\frac{{x}+{y}}{\mathrm{2}}\right)={tana} \\ $$$${x}+{y}=\mathrm{2}{a}=\frac{\mathrm{7}\pi}{\mathrm{6}}=\mathrm{210}^{{o}} \\ $$$$\mathrm{4}{cos}^{\mathrm{2}} \frac{{x}+{y}}{\mathrm{2}}{cos}^{\mathrm{2}} \frac{{x}−{y}}{\mathrm{2}}+\mathrm{4}{sin}^{\mathrm{2}} \frac{{x}+{y}}{\mathrm{2}}{cos}^{\mathrm{2}} \frac{{x}−{y}}{\mathrm{2}}=\mathrm{1} \\ $$$$\mathrm{4}{cos}^{\mathrm{2}} \frac{{x}−{y}}{\mathrm{2}}\left({cos}^{\mathrm{2}} \frac{{x}+{y}}{\mathrm{2}}+{sin}^{\mathrm{2}} \frac{{x}+{y}}{\mathrm{2}}\right)=\mathrm{1} \\ $$$${cos}\frac{{x}−{y}}{\mathrm{2}}=\pm\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${cos}\left(\frac{{x}−{y}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}={cos}\frac{\pi}{\mathrm{3}} \\ $$$${x}−{y}=\mathrm{120}^{{o}} \\ $$$${x}+{y}=\mathrm{210}^{{o}} \\ $$$$\mathrm{2}{x}=\mathrm{330}^{{o}} \:\:\:\:\:{x}=\mathrm{165}^{{o}} \\ $$$${y}=\mathrm{45}^{{o}} \\ $$$${now}\:{if} \\ $$$${cos}\left(\frac{{x}−{y}}{\mathrm{2}}\right)=\frac{−\mathrm{1}}{\mathrm{2}}={cos}\mathrm{120}^{{o}} \\ $$$${x}−{y}=\mathrm{240}^{{o}} \:\:\boldsymbol{{but}}\:\boldsymbol{{look}} \\ $$$$\boldsymbol{{x}}+\boldsymbol{{y}}=\mathrm{210}^{\boldsymbol{{o}}} \:\:\boldsymbol{{but}}\:\boldsymbol{{x}}−\boldsymbol{{y}}=\mathrm{240}^{\boldsymbol{{o}}} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{cos}}\left(\frac{\boldsymbol{{x}}−\boldsymbol{{y}}}{\mathrm{2}}\right)\neq\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$${pls}\:{check}… \\ $$$$ \\ $$
Commented by liki last updated on 27/Mar/20
$$…{Be}\:{blessed}\:{sir},\:{i}\:{understood}\:{well}! \\ $$