Question Number 7039 by Rasheed Soomro last updated on 07/Aug/16
$${Show}\:{without}\:{using}\:{calculator}\:{that} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{sin}\:\mathrm{20sin}\:\mathrm{150sin}\:\mathrm{160}}{\mathrm{sin}\:\mathrm{10}\:\mathrm{sin}\:\mathrm{140}+\mathrm{sin}\:\mathrm{20}\:\mathrm{sin}\:\mathrm{150}\:\mathrm{cos}\:\mathrm{160}}\right)=\mathrm{130} \\ $$
Commented by Yozzii last updated on 08/Aug/16
![All figures are assumed to be in degrees. ((sin20sin150sin160)/(sin10sin(180−40)+sin20sin150cos160)) =((sin20sin150sin160)/(sin10sin40+sin20sin150cos160)) =((sin20sin150sin160)/(2sin10sin20cos20+sin20sin150cos160)) =((sin20sin150sin160)/(sin20(2sin10cos20+sin150cos160))) =((sin150sin160)/(2sin10cos20+sin150cos160)) =((sin150sin160)/(sin30−sin10−sin30cos20)) =((sin150sin160)/(sin30(1−cos20)−sin10)) =((sin150sin160)/(sin30(2sin^2 10)−sin10)) =((sin30sin20)/(sin10(sin10−1))) =((2sin30cos10)/((sin10−1))) =((cos10)/(sin10−1)) ((sin20sin150sin160)/(sin10sin(180−40)+sin20sin150cos160))=((cos10)/(sin10−1)) Now, ((cos10)/(sin10−1))−tan130 =((cos10)/(sin10−1))+tan50 =((cos10)/(sin10−1))+((sin50)/(cos50)) =((cos10cos50+sin10sin50−sin50)/(cos50(sin10−1))) =((cos(50−10)−cos(90−50))/(cos50(sin10−1))) =((cos40−cos40)/(cos50(sin10−1))) =0 ∴((cos10)/(sin10−1))=tan130 ⇒tan^(−1) ((cos10)/(sin10−1))=130 or tan^(−1) [((sin20sin150sin160)/(sin10sin140+sin20sin150cos160))]=130](Q7052.png)
$${All}\:{figures}\:{are}\:{assumed}\:{to}\:{be}\:{in}\:{degrees}. \\ $$$$ \\ $$$$\frac{{sin}\mathrm{20}{sin}\mathrm{150}{sin}\mathrm{160}}{{sin}\mathrm{10}{sin}\left(\mathrm{180}−\mathrm{40}\right)+{sin}\mathrm{20}{sin}\mathrm{150}{cos}\mathrm{160}} \\ $$$$=\frac{{sin}\mathrm{20}{sin}\mathrm{150}{sin}\mathrm{160}}{{sin}\mathrm{10}{sin}\mathrm{40}+{sin}\mathrm{20}{sin}\mathrm{150}{cos}\mathrm{160}} \\ $$$$=\frac{{sin}\mathrm{20}{sin}\mathrm{150}{sin}\mathrm{160}}{\mathrm{2}{sin}\mathrm{10}{sin}\mathrm{20}{cos}\mathrm{20}+{sin}\mathrm{20}{sin}\mathrm{150}{cos}\mathrm{160}} \\ $$$$=\frac{{sin}\mathrm{20}{sin}\mathrm{150}{sin}\mathrm{160}}{{sin}\mathrm{20}\left(\mathrm{2}{sin}\mathrm{10}{cos}\mathrm{20}+{sin}\mathrm{150}{cos}\mathrm{160}\right)} \\ $$$$=\frac{{sin}\mathrm{150}{sin}\mathrm{160}}{\mathrm{2}{sin}\mathrm{10}{cos}\mathrm{20}+{sin}\mathrm{150}{cos}\mathrm{160}} \\ $$$$=\frac{{sin}\mathrm{150}{sin}\mathrm{160}}{{sin}\mathrm{30}−{sin}\mathrm{10}−{sin}\mathrm{30}{cos}\mathrm{20}} \\ $$$$=\frac{{sin}\mathrm{150}{sin}\mathrm{160}}{{sin}\mathrm{30}\left(\mathrm{1}−{cos}\mathrm{20}\right)−{sin}\mathrm{10}} \\ $$$$=\frac{{sin}\mathrm{150}{sin}\mathrm{160}}{{sin}\mathrm{30}\left(\mathrm{2}{sin}^{\mathrm{2}} \mathrm{10}\right)−{sin}\mathrm{10}} \\ $$$$=\frac{{sin}\mathrm{30}{sin}\mathrm{20}}{{sin}\mathrm{10}\left({sin}\mathrm{10}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{2}{sin}\mathrm{30}{cos}\mathrm{10}}{\left({sin}\mathrm{10}−\mathrm{1}\right)} \\ $$$$=\frac{{cos}\mathrm{10}}{{sin}\mathrm{10}−\mathrm{1}} \\ $$$$\frac{{sin}\mathrm{20}{sin}\mathrm{150}{sin}\mathrm{160}}{{sin}\mathrm{10}{sin}\left(\mathrm{180}−\mathrm{40}\right)+{sin}\mathrm{20}{sin}\mathrm{150}{cos}\mathrm{160}}=\frac{{cos}\mathrm{10}}{{sin}\mathrm{10}−\mathrm{1}} \\ $$$$ \\ $$$${Now}, \\ $$$$\frac{{cos}\mathrm{10}}{{sin}\mathrm{10}−\mathrm{1}}−{tan}\mathrm{130} \\ $$$$=\frac{{cos}\mathrm{10}}{{sin}\mathrm{10}−\mathrm{1}}+{tan}\mathrm{50} \\ $$$$=\frac{{cos}\mathrm{10}}{{sin}\mathrm{10}−\mathrm{1}}+\frac{{sin}\mathrm{50}}{{cos}\mathrm{50}} \\ $$$$=\frac{{cos}\mathrm{10}{cos}\mathrm{50}+{sin}\mathrm{10}{sin}\mathrm{50}−{sin}\mathrm{50}}{{cos}\mathrm{50}\left({sin}\mathrm{10}−\mathrm{1}\right)} \\ $$$$=\frac{{cos}\left(\mathrm{50}−\mathrm{10}\right)−{cos}\left(\mathrm{90}−\mathrm{50}\right)}{{cos}\mathrm{50}\left({sin}\mathrm{10}−\mathrm{1}\right)} \\ $$$$=\frac{{cos}\mathrm{40}−{cos}\mathrm{40}}{{cos}\mathrm{50}\left({sin}\mathrm{10}−\mathrm{1}\right)} \\ $$$$=\mathrm{0} \\ $$$$\therefore\frac{{cos}\mathrm{10}}{{sin}\mathrm{10}−\mathrm{1}}={tan}\mathrm{130} \\ $$$$\Rightarrow{tan}^{−\mathrm{1}} \frac{{cos}\mathrm{10}}{{sin}\mathrm{10}−\mathrm{1}}=\mathrm{130} \\ $$$${or} \\ $$$${tan}^{−\mathrm{1}} \left[\frac{{sin}\mathrm{20}{sin}\mathrm{150}{sin}\mathrm{160}}{{sin}\mathrm{10}{sin}\mathrm{140}+{sin}\mathrm{20}{sin}\mathrm{150}{cos}\mathrm{160}}\right]=\mathrm{130} \\ $$$$ \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 08/Aug/16
$${T}\mathcal{H}{an}\Bbbk\mathcal{S}!\:\mathcal{N}{ice}\:{approach}! \\ $$
Answered by Yozzii last updated on 08/Aug/16
$${Check}\:{for}\:{an}\:{answer}\:{in}\:{the}\:{comments}. \\ $$