Question

If $\cos ec\theta - \sin \theta = {a^3}$;$\sec \theta - \cos \theta = {b^3}$ then ${a^2}{b^2}({a^2} + {b^2}) =$
A) 1
B) 2
C) ${\sin ^2}\theta$

Answer 1

This problem involves trigonometric equations with respective relations. For solving this, we require few basic trigonometric formulas such as, $\cos ec\theta = \dfrac{1}{{\sin \theta }}$ , $\sec \theta = \dfrac{1}{{\cos \theta }}$ , ${\sin ^2}\theta + {\cos ^2}\theta = 1$ . Thus using proper trigonometric formulas in the needed time will lead us to the solution.

Complete step-by-step answer:
Step 1: Let us start by simplifying the given equation into a much more simpler form as we can. First let us name the two given relations as Formula 1 and 2.
$\cos ec\theta - \sin \theta = {a^3}$ … Formula 1
$\sec \theta - \cos \theta = {b^3}$ … Formula 2
Step 2: Now using the formula, $\cos ec\theta = \dfrac{1}{{\sin \theta }}$ in Formula 1 we get,
$\dfrac{1}{{\sin \theta }} - \sin \theta = {a^3}$
Simplifying above by taking LCM on RHS, we get
$\dfrac{{1 - {{\sin }^2}\theta }}{{\sin \theta }} = {a^3}$
$\dfrac{{{{\cos }^2}\theta }}{{\sin \theta }} = {a^3}$ Since ${\sin ^2}\theta + {\cos ^2}\theta = 1$ imples $1 - {\sin ^2}\theta = {\cos ^2}\theta$
Solving for a, we get
$a = {\left( {\dfrac{{{{\cos }^2}\theta }}{{\sin \theta }}} \right)^{\dfrac{1}{3}}}$
Step 3: Similarly we solve for ‘b’ by using the formulas $\sec \theta = \dfrac{1}{{\cos \theta }}$ and $1 - {\cos ^2}\theta = {\sin ^2}\theta$
Thus $b = {\left( {\dfrac{{{{\sin }^2}\theta }}{{\cos \theta }}} \right)^{\dfrac{1}{3}}}$
Step 4: We have obtained a and b in terms of sin and cosine functions. Now let us check for what we have to prove, which is ${a^2}{b^2}({a^2} + {b^2})$ Expanding the bracket we get,
${a^2}{b^2}({a^2} + {b^2}) = {a^4}{b^2} + {a^2}{b^4}$
Substituting the values obtained for a and b,
${\left( {\dfrac{{{{\cos }^2}\theta }}{{\sin \theta }}} \right)^{\dfrac{4}{3}}}{\left( {\dfrac{{{{\sin }^2}\theta }}{{\cos \theta }}} \right)^{\dfrac{2}{3}}} + {\left( {\dfrac{{{{\cos }^2}\theta }}{{\sin \theta }}} \right)^{\dfrac{2}{3}}}{\left( {\dfrac{{{{\sin }^2}\theta }}{{\cos \theta }}} \right)^{\dfrac{4}{3}}}$ (Since ${\left( {{x^m}} \right)^n} = {x^{mn}}$ )
$= \dfrac{{{{\left( {{{\cos }^2}\theta } \right)}^{\dfrac{4}{3}}}}}{{{{(\cos \theta )}^{\dfrac{2}{3}}}}}\dfrac{{{{\left( {{{\sin }^2}\theta } \right)}^{\dfrac{2}{3}}}}}{{{{\left( {\sin \theta } \right)}^{\dfrac{4}{3}}}}} + \dfrac{{{{\left( {{{\cos }^2}\theta } \right)}^{\dfrac{2}{3}}}}}{{{{(\cos \theta )}^{\dfrac{4}{3}}}}}\dfrac{{{{\left( {{{\sin }^2}\theta } \right)}^{\dfrac{4}{3}}}}}{{{{\left( {\sin \theta } \right)}^{\dfrac{2}{3}}}}}$
Simplifying the powers of each term,
$= \dfrac{{{{\left( {\cos \theta } \right)}^{\dfrac{8}{3}}}}}{{{{(\cos \theta )}^{\dfrac{2}{3}}}}}\dfrac{{{{\left( {\sin \theta } \right)}^{\dfrac{4}{3}}}}}{{{{\left( {\sin \theta } \right)}^{\dfrac{4}{3}}}}} + \dfrac{{{{\left( {\cos \theta } \right)}^{\dfrac{4}{3}}}}}{{{{(\cos \theta )}^{\dfrac{4}{3}}}}}\dfrac{{{{\left( {\sin \theta } \right)}^{\dfrac{8}{3}}}}}{{{{\left( {\sin \theta } \right)}^{\dfrac{2}{3}}}}}$ (Since ${({\cos ^m}\theta )^n} = {(\cos \theta )^{mn}}$)
$= {(\cos \theta )^{\dfrac{8}{3} - \dfrac{2}{3}}}.{(\sin \theta )^{\dfrac{4}{3} - \dfrac{4}{3}}} + {(\cos \theta )^{\dfrac{4}{3} - \dfrac{4}{3}}}.{(\sin \theta )^{\dfrac{8}{3} - \dfrac{2}{3}}}$
$= {(\cos \theta )^{\dfrac{6}{3}}}.{(\sin \theta )^0} + {(\cos \theta )^0}.{(\sin \theta )^{\dfrac{6}{3}}}$
As anything power zero is one, we obtain the above relation as,
${\cos ^2}\theta + {\sin ^2}\theta = 1$
Thus ${a^2}{b^2}({a^2} + {b^2}) = 1$ which is option 1.
Final answer:
${a^2}{b^2}({a^2} + {b^2}) =1$

Thus option 1 is correct.

Note: Usage of needed basic trigonometric formulas is the important factor in solving similar questions. Power simplification is an error prone step which may lead easily to wrong conclusions. Also we can use other two basic trigonometric relations: $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }},\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ . As they have provided the trigonometric values of both ${a^3}$and ${b^3}$ , we can substitute this directly to the expansion of ${a^2}{b^2}({a^2} + {b^2})$.