Question

Which of the following options is/are equal to $ta{n}^{2}A+co{t}^{2}A+2$ ?

• A. $\frac{1}{co{s}^{2}Asi{n}^{2}A}$
• B. $se{c}^{2}Acose{c}^{2}A$
• C. $ta{n}^{2}Aco{t}^{2}A$
• D. $se{c}^{2}A+cose{c}^{2}A$

Answer 1

Answer: (A), (B), (D)

$se{c}^{2}A+cose{c}^{2}A$

$ta{n}^{2}A+co{t}^{2}A+2\text{can be written as}(1+ta{n}^{2}A)+(1+co{t}^{2}A)$

We know that, $1+ta{n}^{2}A=se{c}^{2}A\text{and}1+co{t}^{2}A=cose{c}^{2}A$

$\Rightarrow ta{n}^{2}A+co{t}^{2}A+2=se{c}^{2}A+cose{c}^{2}A$

$=\frac{1}{co{s}^{2}A}+\frac{1}{si{n}^{2}A}$

$=\frac{si{n}^{2}A+co{s}^{2}A}{co{s}^{2}Asi{n}^{2}A}...(si{n}^{2}A+co{s}^{2}A=1)$

$=\frac{1}{co{s}^{2}Asi{n}^{2}A}$

$=se{c}^{2}Acose{c}^{2}A$