Question : If $\sec\theta+\tan\theta=2$, then the value of $\sec\theta$ is:
Option 1: $\frac{4}{5}$
Option 2: $5$
Option 3: $\frac{5}{4}$
Option 4: $\sqrt{2}$
Correct Answer: $\frac{5}{4}$
Solution : Given: $\sec\theta+\tan\theta=2$ -----(1) We know, $\sec^2\theta-\tan^2\theta=1$ Or, $(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1$ $\therefore (\sec\theta-\tan\theta)=\frac{1}{2}$ ------(2) Solving equations 1 and 2, we get, $2\sec\theta=2 +\frac{1}{2}=\frac{5}{2}$ $\therefore \sec\theta=\frac{5}{4}$ Hence, the correct answer is $\frac{5}{4}$.
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Question : If $\frac{1}{\operatorname{cosec} \theta+1}+\frac{1}{\operatorname{cosec} \theta-1}=2 \sec \theta, 0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{\tan \theta+2 \sec \theta}{\operatorname{cosec} \theta}$ is:
Question : If $\cos\theta=\frac{3}{5}$, then the value of $\sin\theta.\sec\theta.\tan\theta$ is:
Question : If $4-2 \sin ^2 \theta-5 \cos \theta=0,0^{\circ}<\theta<90^{\circ}$, then the value of $\cos \theta-\tan \theta$ is:
Question : If $4-2 \sin ^2 \theta-5 \cos \theta=0,0^{\circ}<\theta<90^{\circ}$, then the value of $\cos \theta+\tan \theta$ is:
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