Menu Close

Question-205001

Tinku Tara 0 Comments
Pin




Question Number 205001 by rajusasmal last updated on 05/Mar/24
Answered by TonyCWX08 last updated on 05/Mar/24
44. x^2 −y^2 =(asec(θ)+btan(θ))^2 −(atan(θ)+bsec(θ))^2 =a^2 sec^2 (θ)+2absec(θ)tan(θ)+b^2 tan^2 (θ)−a^2 tan^2 (θ)−2abtan(θ)sec(θ)−b^2 sec^2 (θ) =a^2 sec^2 (θ)−a^2 tan^2 (θ)+b^2 tan^2 (θ)−b^2 sec^2 (θ) =a^2 (sec^2 (θ)−tan^2 (θ))+b^2 (tan^2 (θ)−sec^2 (θ)) =a^2 (1)+b^2 (−1) =a^2 −b^2 45. 3sin(x)+5cos(x)=5 3sin^2 (x)+30sin(x)cos(x)+25cos^2 (x)=25 30sin(x)cos(x)=25−3sin^2 (x)−25cos^2 (x) (5sin(x)−3cos(x))^2 =25sin^2 (x)−30sin(x)cos(x)+9cos^2 (x) =25sin^2 (x)−(25−3sin^2 (x)−25cos^2 (x))+9cos^2 (x) =25sin^2 (x)−25+3sin^2 (x)+25cos^2 (x)+9cos^2 (x) =34sin^2 (x)+34cos^2 (x)−25 =34(sin^2 (x)+cos^2 (x))−25 =34−25 =9 5sin(x)−3cos(x)=±(√9)=±3
$$\mathrm{44}.\: \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} \\ $$$$=\left({asec}\left(\theta\right)+{btan}\left(\theta\right)\right)^{\mathrm{2}} −\left({atan}\left(\theta\right)+{bsec}\left(\theta\right)\right)^{\mathrm{2}} \\ $$$$={a}^{\mathrm{2}} {sec}^{\mathrm{2}} \left(\theta\right)+\mathrm{2}{absec}\left(\theta\right){tan}\left(\theta\right)+{b}^{\mathrm{2}} {tan}^{\mathrm{2}} \left(\theta\right)−{a}^{\mathrm{2}} {tan}^{\mathrm{2}} \left(\theta\right)−\mathrm{2}{abtan}\left(\theta\right){sec}\left(\theta\right)−{b}^{\mathrm{2}} {sec}^{\mathrm{2}} \left(\theta\right) \\ $$$$={a}^{\mathrm{2}} {sec}^{\mathrm{2}} \left(\theta\right)−{a}^{\mathrm{2}} {tan}^{\mathrm{2}} \left(\theta\right)+{b}^{\mathrm{2}} {tan}^{\mathrm{2}} \left(\theta\right)−{b}^{\mathrm{2}} {sec}^{\mathrm{2}} \left(\theta\right) \\ $$$$={a}^{\mathrm{2}} \left({sec}^{\mathrm{2}} \left(\theta\right)−{tan}^{\mathrm{2}} \left(\theta\right)\right)+{b}^{\mathrm{2}} \left({tan}^{\mathrm{2}} \left(\theta\right)−{sec}^{\mathrm{2}} \left(\theta\right)\right) \\ $$$$={a}^{\mathrm{2}} \left(\mathrm{1}\right)+{b}^{\mathrm{2}} \left(−\mathrm{1}\right) \\ $$$$={a}^{\mathrm{2}} −{b}^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{45}. \\ $$$$\mathrm{3}{sin}\left({x}\right)+\mathrm{5}{cos}\left({x}\right)=\mathrm{5} \\ $$$$\mathrm{3}{sin}^{\mathrm{2}} \left({x}\right)+\mathrm{30}{sin}\left({x}\right){cos}\left({x}\right)+\mathrm{25}{cos}^{\mathrm{2}} \left({x}\right)=\mathrm{25} \\ $$$$\mathrm{30}{sin}\left({x}\right){cos}\left({x}\right)=\mathrm{25}−\mathrm{3}{sin}^{\mathrm{2}} \left({x}\right)−\mathrm{25}{cos}^{\mathrm{2}} \left({x}\right) \\ $$$$ \\ $$$$\left(\mathrm{5}{sin}\left({x}\right)−\mathrm{3}{cos}\left({x}\right)\right)^{\mathrm{2}} \\ $$$$=\mathrm{25}{sin}^{\mathrm{2}} \left({x}\right)−\mathrm{30}{sin}\left({x}\right){cos}\left({x}\right)+\mathrm{9}{cos}^{\mathrm{2}} \left({x}\right) \\ $$$$=\mathrm{25}{sin}^{\mathrm{2}} \left({x}\right)−\left(\mathrm{25}−\mathrm{3}{sin}^{\mathrm{2}} \left({x}\right)−\mathrm{25}{cos}^{\mathrm{2}} \left({x}\right)\right)+\mathrm{9}{cos}^{\mathrm{2}} \left({x}\right) \\ $$$$=\mathrm{25}{sin}^{\mathrm{2}} \left({x}\right)−\mathrm{25}+\mathrm{3}{sin}^{\mathrm{2}} \left({x}\right)+\mathrm{25}{cos}^{\mathrm{2}} \left({x}\right)+\mathrm{9}{cos}^{\mathrm{2}} \left({x}\right) \\ $$$$=\mathrm{34}{sin}^{\mathrm{2}} \left({x}\right)+\mathrm{34}{cos}^{\mathrm{2}} \left({x}\right)−\mathrm{25} \\ $$$$=\mathrm{34}\left({sin}^{\mathrm{2}} \left({x}\right)+{cos}^{\mathrm{2}} \left({x}\right)\right)−\mathrm{25} \\ $$$$=\mathrm{34}−\mathrm{25} \\ $$$$=\mathrm{9} \\ $$$$ \\ $$$$\mathrm{5}{sin}\left({x}\right)−\mathrm{3}{cos}\left({x}\right)=\pm\sqrt{\mathrm{9}}=\pm\mathrm{3} \\ $$$$ \\ $$
Answered by TonyCWX08 last updated on 05/Mar/24
36. x^2 +y^2 +z^2 =r^2 sin^2 (A)cos^2 (C)+r^2 sin^2 (A)sin^2 (C)+r^2 cos^2 (A) =r^2 (sin^2 (A)(sin^2 (C)+cos^2 (C))+cos^2 (A)) =r^2 (sin^2 (A)+cos^2 (A)) =r^2 (1) =r^2 40. sin(x)+sin^2 (x)=1 sin^2 (x)=1−sin(x) 1−cos^2 (x)=1−sin(x) −cos^2 (x)=−sin(x) cos^2 (x)=sin(x) cos^4 (x)=sin^2 (x) cos^2 (x)+cos^4 (x) =sin(x)+sin^2 (x)=1
$$\mathrm{36}. \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \\ $$$$={r}^{\mathrm{2}} {sin}^{\mathrm{2}} \left({A}\right){cos}^{\mathrm{2}} \left({C}\right)+{r}^{\mathrm{2}} {sin}^{\mathrm{2}} \left({A}\right){sin}^{\mathrm{2}} \left({C}\right)+{r}^{\mathrm{2}} {cos}^{\mathrm{2}} \left({A}\right) \\ $$$$={r}^{\mathrm{2}} \left({sin}^{\mathrm{2}} \left({A}\right)\left({sin}^{\mathrm{2}} \left({C}\right)+{cos}^{\mathrm{2}} \left({C}\right)\right)+{cos}^{\mathrm{2}} \left({A}\right)\right) \\ $$$$={r}^{\mathrm{2}} \left({sin}^{\mathrm{2}} \left({A}\right)+{cos}^{\mathrm{2}} \left({A}\right)\right) \\ $$$$={r}^{\mathrm{2}} \left(\mathrm{1}\right) \\ $$$$={r}^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{40}. \\ $$$${sin}\left({x}\right)+{sin}^{\mathrm{2}} \left({x}\right)=\mathrm{1} \\ $$$${sin}^{\mathrm{2}} \left({x}\right)=\mathrm{1}−{sin}\left({x}\right) \\ $$$$\mathrm{1}−{cos}^{\mathrm{2}} \left({x}\right)=\mathrm{1}−{sin}\left({x}\right) \\ $$$$−{cos}^{\mathrm{2}} \left({x}\right)=−{sin}\left({x}\right) \\ $$$${cos}^{\mathrm{2}} \left({x}\right)={sin}\left({x}\right) \\ $$$${cos}^{\mathrm{4}} \left({x}\right)={sin}^{\mathrm{2}} \left({x}\right) \\ $$$${cos}^{\mathrm{2}} \left({x}\right)+{cos}^{\mathrm{4}} \left({x}\right) \\ $$$$={sin}\left({x}\right)+{sin}^{\mathrm{2}} \left({x}\right)=\mathrm{1} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *