Question : $\frac{2\sin68°}{\cos22°}-\frac{2\cot15°}{5\tan75°} -\frac{3\tan45°\cdot\ \tan20°\cdot\ \tan40°\cdot\ \tan50°\cdot\ \tan70°}{5}$ is equal to:
Option 1: –1
Option 2: 0
Option 3: 1
Option 4: 2
New: SSC CHSL tier 1 answer key 2024 out | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: 1
Solution : $\frac{2\sin68°}{\cos22°}-\frac{2\cot15°}{5\tan75°} - \: \frac{3\tan45°.\tan20°.\tan40°.\tan50°.\tan70°}{5}$ = $\frac{2\sin68°}{\cos(90°-68°)}-\frac{2\cot(90°-75°)}{5\tan75°} - \frac{3\cdot 1\cdot\tan20°\cdot\tan40°\cdot\tan(90°-40°)\cdot\tan(90°-20°)}{5}$ = $\frac{2\sin68°}{\sin68°}-\frac{2\tan75°}{5\tan75°} -\frac{3\cdot 1\cdot\tan20°\cdot\tan40°\cdot\cot40°\cdot\cot20°}{5}$ = $2×1 - \frac{2}{5}× 1 - \frac{3}{5}$ = $2-\frac{2}{5}-\frac{3}{5}$ = $\frac{10-2-3}{5}$ = $1$ Hence, the correct answer is 1.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Question : The value of $\tan10°\tan15°\tan75°\tan80°$ is:
Question : The value of the following is: $\frac{\left (\tan20^{\circ} \right )^{2}}{\left (\operatorname{cosec70^{\circ}} \right )^{2}}+\frac{\left (\cot20^{\circ} \right )^{2}}{\left (\sec70^{\circ} \right )^{2}}+2\tan15^{\circ} \tan45^{\circ} \tan75^{\circ}$
Question : The value of $\frac{\cos65^\circ}{\sin25^\circ}+\frac{\cos55^\circ}{\sin35^\circ}-\tan50^\circ\cot50^\circ$ is equal to:
Question : The value of $\cos \ 0°+\cos\ 1°+\cos\ 2°......\cos\ 180°$ is:
Question : If $xy+yz+zx=0$, then $(\frac{1}{x^2–yz}+\frac{1}{y^2–zx}+\frac{1}{z^2–xy})$$(x,y,z \neq 0)$ is equal to:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile