Question
Which one of the following options is correct ?
A.
$${\sin ^2}{30^ \circ },{\sin ^2}{45^ \circ },{\sin ^2}{60^ \circ }{\text{are in G}}{\text{.P}}{\text{.}}$$
B.
$${\cos ^2}{30^ \circ },{\cos ^2}{45^ \circ },{\cos ^2}{60^ \circ }{\text{are in G}}{\text{.P}}{\text{.}}$$
C.
$${\cot ^2}{30^ \circ },{\cot ^2}{45^ \circ },{\cot ^2}{60^ \circ }{\text{are in G}}{\text{.P}}{\text{.}}$$
D.
$${\tan ^2}{30^ \circ },{\tan ^2}{45^ \circ },{\tan ^2}{60^ \circ }{\text{are in G}}{\text{.P}}{\text{.}}$$
Answer :
$${\tan ^2}{30^ \circ },{\tan ^2}{45^ \circ },{\tan ^2}{60^ \circ }{\text{are in G}}{\text{.P}}{\text{.}}$$
Solution :
Three numbers $$a, b$$ and $$c$$ will be in G.P. if $$b^2 = ac.$$ Only option $$(D)$$ i.e. $${\tan ^2}{30^ \circ },{\tan^2}{45^ \circ }{\text{and}}\,{\tan^2}{60^ \circ }{\text{are in G}}{\text{.P}}{\text{.}}$$
$$\eqalign{
& \because {\text{ta}}{{\text{n}}^2}{30^ \circ } = \frac{1}{3} \cr
& {\tan ^2}{45^ \circ } = 1 \cr
& {\text{and }}{\tan ^2}{60^ \circ } = 3 \cr
& \therefore {\tan ^2}{30^ \circ },{\tan^2}{45^ \circ }{\text{and}}\,{\tan^2}{60^ \circ }{\text{are in G}}{\text{.P}}{\text{.}} \cr} $$