If ∠A and ∠B are complementary to each other, then the value of sec2A + sec2B - sec2A.sec2B is?
A. 1
B. -1
C. 2
D. 0
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & {\text{A + B}} = {90^ \circ } \cr & {\text{B}} = {90^ \circ } - {\text{A}} \cr & {\sec ^2}A + se{c^2}{\text{B}} - {\text{se}}{{\text{c}}^2}{\text{A}}{\text{.se}}{{\text{c}}^2}{\text{B}} \cr} $$$$ = {\sec ^2}A + se{c^2}\left( {{{90}^ \circ } - {\text{A}}} \right) - $$ $${\text{se}}{{\text{c}}^2}{\text{A}}{\text{.se}}{{\text{c}}^2}$$ $$\left( {{{90}^ \circ } - {\text{A}}} \right)$$
$$\eqalign{ & = {\sec ^2}A + {\operatorname{cosec} ^2}{\text{A}} - {\text{se}}{{\text{c}}^2}{\text{A}}{\text{.cose}}{{\text{c}}^2}{\text{A}} \cr & = \frac{1}{{{{\cos }^2}{\text{A}}}} + \frac{1}{{si{n^2}{\text{A}}}} - \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}}} \times \frac{1}{{{\text{si}}{{\text{n}}^2}{\text{A}}}} \cr & = \frac{{{{\sin }^2}{\text{A + co}}{{\text{s}}^2}{\text{A}}}}{{{\text{co}}{{\text{s}}^2}{\text{A}}{\text{.}}{{\sin }^2}{\text{A}}}} - \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}{\text{.}}{{\sin }^2}{\text{A}}}} \cr & = \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}.{{\sin }^2}{\text{A}}}} - \frac{1}{{{\text{co}}{{\text{s}}^2}{\text{A}}.{{\sin }^2}{\text{A}}}} \cr & = 0 \cr} $$
Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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