Question
State True or False for the following statement:
If $\text{cosec x }=1+\cot\text{x}$ then $\text{x}=2\text{n}\pi,2\text{n}\pi+\frac{\pi}{2}$

Answer

True. Solution: Given that, $\text{cosec}\text{x}=1+\cot\text{x}$ $\Rightarrow\frac{1}{\sin\text{x}}=1+\frac{\cos\text{x}}{\sin\text{x}}$ $\Rightarrow\frac{1}{\sin\text{x}}=\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}}$ $\Rightarrow\sin\text{x}+\cos\text{x}=1$ $\Rightarrow\frac{1}{\sqrt2}\sin\text{x}+\frac{1}{\sqrt{2}}\cos\text{x}=\frac{1}{\sqrt2}$ $\Rightarrow\sin\frac{\pi}{4}\sin\text{x}+\cos\frac{\pi}{4}\cos\text{x}=\frac{1}{\sqrt2}$ $\Rightarrow\cos\Big(\text{x}-\frac{\pi}{4}\Big)=\frac{1}{\sqrt2}$ $\Rightarrow\cos\Big(\text{x}-\frac{\pi}{4}\Big)=\cos\frac{\pi}{4}$$\text{x}=2\text{n}\pi+\frac{\pi}{4}+\frac{\pi}{4}\Rightarrow\text{x}=2\text{n}\pi+\frac{\pi}{2}$
or $\text{x}=2\text{n}\pi+\frac{\pi}{4}-\frac{\pi}{4}\Rightarrow\text{x}=2\text{n}\pi$ Hence, the given statement is 'True'.

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