Question

$(\cos^{2}0^{\circ}+\sin^{2}30^{\circ})(\tan^{2}45^{\circ}+\cot^{2}60^{\circ})(\cos^{2}180^{\circ}+\sin^{2}90^{\circ})=$

• A. $\frac{16}{3}$
• B. $\frac{14}{3}$
• C. $\frac{12}{3}$
• D. $\frac{10}{3}$

Hint:

Memorizing standard trigonometric values helps solve such questions in a few seconds.

Answer 1

The Correct Option is D

Concept: Substitute the standard trigonometric values and simplify each bracket separately.

Step 1: Evaluate the first bracket.
\[ \cos0^\circ =1,\qquad \sin30^\circ=\frac12 \] \[ \cos^20^\circ+\sin^230^\circ =1+\frac14 =\frac54 \]

Step 2: Evaluate the second bracket.
\[ \tan45^\circ=1,\qquad \cot60^\circ=\frac1{\sqrt3} \] \[ \tan^245^\circ+\cot^260^\circ =1+\frac13 =\frac43 \]

Step 3: Evaluate the third bracket.
\[ \cos180^\circ=-1,\qquad \sin90^\circ=1 \] \[ \cos^2180^\circ+\sin^290^\circ =1+1 =2 \]

Step 4: Multiply all factors.
\[ \frac54 \times \frac43 \times 2 =\frac{10}{3} \]