Question 52

If $$\sin \theta$$ = 1/$$\surd 2$$ then $$(\tan \theta + \cos \theta)$$ =

Solution

$$\sin \theta$$ = 1/$$\surd 2$$

$$\sin\theta\ =\frac{P}{H}=\frac{1}{\sqrt{\ 2}}$$

$$B^2 =H^2-P^2$$

$$B^2=2\ -\ 1\ =1$$

B = 1

$$\tan\ =\frac{P}{B},\ \cos\ =\frac{B}{H}$$

$$(\tan \theta + \cos \theta)$$ = $$1+\frac{1}{\sqrt{\ 2}}=\frac{\sqrt{\ 2}+1}{\sqrt{\ 2}}$$.

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