Step-by-step explanation:

no solution is possible for the given equation

Verified answer

Appropriate Question: Prove that

[tex]\sf \: \dfrac{ {cos}^{4}q - {sin}^{4}q}{1 + tanq} = \dfrac{cotq - 1}{secq \: cosecq} \\ \\ [/tex]

[tex]\large\underline{\sf{Solution-}}[/tex]

Consider,

[tex]\sf \: \dfrac{ {cos}^{4}q - {sin}^{4}q}{1 + tanq} \\ \\ [/tex]

We know,

[tex]\boxed{\begin{aligned}& \qquad \:\sf \: {x}^{4} - {y}^{4} = (x - y)(x + y)( {x}^{2} - {y}^{2}) \qquad \: \\ \\& \qquad \:\sf \: tanx= \dfrac{sinx}{cosx} \end{aligned}} \qquad \: \\ \\ [/tex]

So, using these results, the above expression can be rewritten as

[tex]\sf \: = \: \dfrac{(cosq - sinq)(cosq + sinq)( {cos}^{2}q + {sin}^{2}q)}{1 + \dfrac{ {sin}q}{ {cos}q} } \\ \\ [/tex]

[tex]\sf \: = \: \dfrac{(cosq - sinq)(cosq + sinq)(1)}{ \dfrac{cosq + {sin}q}{ {cos}q} } \\ \\ [/tex]

[tex]\sf \: = \: cosq(cosq - sinq) \\ \\ [/tex]

[tex]\sf \: = \: cosq \: sinq\left(\dfrac{cosq}{sinq} - 1\right) \\ \\ [/tex]

[tex]\sf \: = \: cosq \: sinq\left(cotq - 1\right) \\ \\ [/tex]

[tex]\sf \: = \: \dfrac{1}{secq} \times \dfrac{1}{cosecq} \times \left(cotq - 1\right) \\ \\ [/tex]

[tex]\sf \: = \: \dfrac{cotq - 1}{secq \: cosecq} \\ \\ [/tex]

Hence,

[tex]\implies\sf \: \boxed{\sf \: \sf \: \dfrac{ {cos}^{4}q - {sin}^{4}q}{1 + tanq} = \dfrac{cotq - 1}{secq \: cosecq} \: \: } \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\begin{gathered} \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sinx = \dfrac{1}{cosecx} }\\ \\ \bigstar \: \bf{cosx = \dfrac{1}{secx} }\\ \\ \bigstar \: \bf{tanx = \dfrac{sinx}{cosx} = \dfrac{1}{cotx} }\\ \\ \bigstar \: \bf{cot x= \dfrac{cosx}{sinx} = \dfrac{1}{tanx} }\\ \\ \bigstar \: \bf{cosec x = \dfrac{1}{sinx} }\\ \\ \bigstar \: \bf{secx = \dfrac{1}{cosx} }\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 } \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\\: \end{array} }}\end{gathered}\end{gathered}\end{gathered} \\ \\[/tex]