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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. 5(a) & c

$$..{plz}\:{i}\:{need}\:{help}\:{qn}\:{no}.\:\mathrm{5}\left({a}\right)\:\&\:{c} \\ $$

Answered by TANMAY PANACEA. last updated on 27/Mar/20

((tan145−tan125)/(1+tan145tan125)) =tan(145−125) =tan20 =cot70 =(1/(tan70)) =((1−tan^2 35)/(2tan35)) =((1−x^2 )/(2x))

$$\frac{{tan}\mathrm{145}−{tan}\mathrm{125}}{\mathrm{1}+{tan}\mathrm{145}{tan}\mathrm{125}} \\ $$$$={tan}\left(\mathrm{145}−\mathrm{125}\right) \\ $$$$={tan}\mathrm{20} \\ $$$$={cot}\mathrm{70} \\ $$$$=\frac{\mathrm{1}}{{tan}\mathrm{70}} \\ $$$$=\frac{\mathrm{1}−{tan}^{\mathrm{2}} \mathrm{35}}{\mathrm{2}{tan}\mathrm{35}} \\ $$$$=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{2}{x}} \\ $$

Commented by liki last updated on 27/Mar/20

....Thank you so much

$$....{Thank}\:{you}\:{so}\:{much} \\ $$

Answered by TANMAY PANACEA. last updated on 27/Mar/20

cosx+cosy=cosa sinx+siny=sina a=((7π)/(12))=105^o from 1) 2cos((x+y)/2)cos((x−y)/2)=cosa from 2) 2sin((x+y)/2)cos((x−y)/2)=sina tan(((x+y)/2))=tana x+y=2a=((7π)/6)=210^o 4cos^2 ((x+y)/2)cos^2 ((x−y)/2)+4sin^2 ((x+y)/2)cos^2 ((x−y)/2)=1 4cos^2 ((x−y)/2)(cos^2 ((x+y)/2)+sin^2 ((x+y)/2))=1 cos((x−y)/2)=±(1/2) cos(((x−y)/2))=(1/2)=cos(π/3) x−y=120^o x+y=210^o 2x=330^o x=165^o y=45^o now if cos(((x−y)/2))=((−1)/2)=cos120^o x−y=240^o but look x+y=210^o but x−y=240^o so cos(((x−y)/2))≠((−1)/2) pls check...

$${cosx}+{cosy}={cosa} \\ $$$${sinx}+{siny}={sina} \\ $$$${a}=\frac{\mathrm{7}\pi}{\mathrm{12}}=\mathrm{105}^{{o}} \\ $$$$\left.{from}\:\mathrm{1}\right)\:\:\mathrm{2}{cos}\frac{{x}+{y}}{\mathrm{2}}{cos}\frac{{x}−{y}}{\mathrm{2}}={cosa} \\ $$$$\left.{from}\:\mathrm{2}\right)\:\mathrm{2}{sin}\frac{{x}+{y}}{\mathrm{2}}{cos}\frac{{x}−{y}}{\mathrm{2}}={sina} \\ $$$${tan}\left(\frac{{x}+{y}}{\mathrm{2}}\right)={tana} \\ $$$${x}+{y}=\mathrm{2}{a}=\frac{\mathrm{7}\pi}{\mathrm{6}}=\mathrm{210}^{{o}} \\ $$$$\mathrm{4}{cos}^{\mathrm{2}} \frac{{x}+{y}}{\mathrm{2}}{cos}^{\mathrm{2}} \frac{{x}−{y}}{\mathrm{2}}+\mathrm{4}{sin}^{\mathrm{2}} \frac{{x}+{y}}{\mathrm{2}}{cos}^{\mathrm{2}} \frac{{x}−{y}}{\mathrm{2}}=\mathrm{1} \\ $$$$\mathrm{4}{cos}^{\mathrm{2}} \frac{{x}−{y}}{\mathrm{2}}\left({cos}^{\mathrm{2}} \frac{{x}+{y}}{\mathrm{2}}+{sin}^{\mathrm{2}} \frac{{x}+{y}}{\mathrm{2}}\right)=\mathrm{1} \\ $$$${cos}\frac{{x}−{y}}{\mathrm{2}}=\pm\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${cos}\left(\frac{{x}−{y}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}={cos}\frac{\pi}{\mathrm{3}} \\ $$$${x}−{y}=\mathrm{120}^{{o}} \\ $$$${x}+{y}=\mathrm{210}^{{o}} \\ $$$$\mathrm{2}{x}=\mathrm{330}^{{o}} \:\:\:\:\:{x}=\mathrm{165}^{{o}} \\ $$$${y}=\mathrm{45}^{{o}} \\ $$$${now}\:{if} \\ $$$${cos}\left(\frac{{x}−{y}}{\mathrm{2}}\right)=\frac{−\mathrm{1}}{\mathrm{2}}={cos}\mathrm{120}^{{o}} \\ $$$${x}−{y}=\mathrm{240}^{{o}} \:\:\boldsymbol{{but}}\:\boldsymbol{{look}} \\ $$$$\boldsymbol{{x}}+\boldsymbol{{y}}=\mathrm{210}^{\boldsymbol{{o}}} \:\:\boldsymbol{{but}}\:\boldsymbol{{x}}−\boldsymbol{{y}}=\mathrm{240}^{\boldsymbol{{o}}} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{cos}}\left(\frac{\boldsymbol{{x}}−\boldsymbol{{y}}}{\mathrm{2}}\right)\neq\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$${pls}\:{check}... \\ $$$$ \\ $$

Commented by liki last updated on 27/Mar/20

...Be blessed sir, i understood well!

$$...{Be}\:{blessed}\:{sir},\:{i}\:{understood}\:{well}! \\ $$

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