Question

$(\sec \theta -\cos \theta {)}^2+(cosec\theta -\sin \theta {)}^2-(\cot \theta -\tan \theta {)}^2=$

[1 mark]

• A. -1
• B. 1
• C. 0
• D. 2

Answer 1

The correct option is B 1

$(\sec \theta -\cos \theta {)}^2={\sec }^2\theta +{\cos }^2\theta -2(\sec \theta \left)\right(\cos \theta \left)\right(cossec\theta -\sin \theta {)}^2=cose{c}^2\theta +{\sin }^2\theta -2\left(cosec\theta \right)\left(\sin \theta \right)(\cot \theta -\tan \theta {)}^2={\cot }^2\theta +{\tan }^2\theta -2(\cot \theta \left)\right(\tan \theta \left)\right(\sec \theta -\cos \theta {)}^2+(cosec\theta -\sin \theta {)}^2-(\cot \theta -\tan \theta {)}^2=({\sec }^2\theta +{\cos }^2\theta -2)+(cose{c}^2\theta +{\sin }^2\theta -2)-({\cot }^2\theta +{\tan }^2\theta -2)$

In the RHS of the equation,

$\left(\sec \theta \right)\left(\cos \theta \right)=1$, $\left(cosec\theta \right)\left(\sin \theta \right)=1$ and $\left(\cot \theta \right)\left(\tan \theta \right)=1$

$(\sec \theta -\cos \theta {)}^2+(cosec\theta -\sin \theta {)}^2-(\cot \theta -\tan \theta {)}^2=({\sec }^2\theta -{\tan }^2\theta )+({\sin }^2\theta +{\cos }^2\theta )+(cose{c}^2\theta -\cot 2\theta )-2-2+2$

$\because ({\sec }^2\theta -{\tan }^2\theta )=1,({\sin }^2\theta +{\cos }^2\theta )=1and(cose{c}^2\theta -{\cot }^2\theta )=1$

$\Rightarrow (\sec \theta -\cos \theta {)}^2+(cosec\theta -\sin \theta {)}^2-(\cot \theta -\tan \theta {)}^2=1+1+1-2$

$\therefore (\sec \theta -\cos \theta {)}^2+(cosec\theta -\sin \theta {)}^2-(\cot \theta -\tan \theta {)}^2=1$