CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

If $$x=\csc{\theta}-\sin{\theta}$$ and $$y=\sec{\theta}-\cos{\theta}$$,  then the value of $${x}^{2}+{y}^{2}$$  $$=cosec^2\theta.sec^2\theta-p$$ . Then $$p=$$


Solution

Given:
$$x=cosec\theta-sin\theta$$ and

$$y=sec\theta-cos\theta$$

Hence,

$$x^2+y^2=cosec^2\theta+sin^2\theta-2cosec\theta sin\theta+sec^2\theta+cos^2\theta-2sec\theta cos\theta$$

$$=1+1+tan^2\theta+1+cot^2\theta-2-2=\dfrac{sin^2\theta}{cos^2\theta}+\dfrac{cos^2\theta}{sin^2\theta}-1=cosec^2\theta sec^2\theta-1$$...................[since, $$sin^2A+cos^2A=1;\ 1+tan^2A=sec^2A;\ 1+cot^2A=cosec^2A]$$

Hence,

$$P=1$$

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image