Question

Prove that $\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }\ =\ 1+\sec \theta \csc \theta$by using formula and identities.

Answer 1

Hint: First consider left hand side of equation and then convert $\tan \theta$ as $\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta$ as $\dfrac{\cos \theta }{\sin \theta }$ and then do the further calculations to get the Right hand side of equation.

Complete step-by-step answer:
In the question we have to prove that
$\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }\ =\ 1+\sec \theta \csc \theta$
Now, let’s consider the left had side of the equation we see that,
$\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }$ …………………………………………………………………………..(i)
Now we will use formula that is$\tan \theta \ =\ \dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta \ =\ \dfrac{\cos \theta }{\sin \theta }$ , then substitute it in expression (i) so we get,

& \dfrac{\dfrac{\sin \theta }{\cos \theta }}{1-\dfrac{\cos \theta }{\sin \theta }}+\dfrac{\dfrac{\cos \theta }{\sin \theta }}{1-\dfrac{\sin \theta }{\cos \theta }} \\\ & \Rightarrow \dfrac{\dfrac{\sin \theta }{\cos \theta }}{\dfrac{sin\theta -\cos \theta }{\sin \theta }}+\dfrac{\dfrac{\cos \theta }{\sin \theta }}{\dfrac{cos\theta -\sin \theta }{\cos \theta }} \\\ & \Rightarrow \dfrac{{{\sin }^{2}}\theta }{cos\theta (\sin \theta -\cos \theta )}+\dfrac{{{\cos }^{2}}\theta }{\sin \theta (\cos \theta -\sin \theta )} \\\ & \Rightarrow \dfrac{{{\sin }^{2}}\theta }{\sin \theta \cos \theta -{{\cos }^{2}}\theta }+\dfrac{{{\cos }^{2}}\theta }{\sin \theta \cos \theta -{{\sin }^{2}}\theta }\ \\\ \end{aligned}$$ Now we take LCM of the term so we get, $$\dfrac{{{\sin }^{2}}\theta \left( \sin \theta \cos \theta -{{\sin }^{2}}\theta \right)+{{\cos }^{2}}\theta \left( \sin \theta \cos \theta -{{\cos }^{2}}\theta \right)}{\left( \sin \theta \cos \theta -{{\cos }^{2}}\theta \right)\left( \sin \theta \cos \theta -{{\sin }^{2}}\theta \right)}$$ So, it can be further simplified as, $$\dfrac{{{\sin }^{3}}\theta \cos \theta -{{\sin }^{4}}\theta +\sin \theta {{\cos }^{3}}\theta -{{\cos }^{4}}\theta }{\cos \theta \left( \sin \theta -\cos \theta \right)\sin \theta \left( \cos \theta -\sin \theta \right)}$$ Which can write it as, $$\dfrac{\sin \theta \cos \theta \left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)-\left( {{\sin }^{4}}\theta +{{\cos }^{4}}\theta \right)}{-\sin \theta \cos \theta {{\left( \sin \theta -\cos \theta \right)}^{2}}}$$ Now, we will change $$\left( {{\sin }^{4}}\theta +{{\cos }^{4}}\theta \right)$$ as $$\left\\{ \left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right\\}$$ and we will use identity $${{\sin }^{2}}\theta +{{\cos }^{2}}\theta \ =\ 1$$. Hence, it can be written as, $$\dfrac{\sin \theta \cos \theta -\left\\{ 1-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right\\}}{-\sin \theta \cos \theta {{\left( \sin \theta -\cos \theta \right)}^{2}}}$$ Now, we will multiply -1 to both numerator and denominator so we get, $$\dfrac{1-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta -\sin \theta \cos \theta }{\sin \theta \cos \theta {{\left( \sin \theta -\cos \theta \right)}^{2}}}$$ Now, we will factorize the numerator so we will get, $$\dfrac{1-2\sin \theta \cos \theta +\sin \theta \cos \theta -2{{\sin }^{2}}\theta {{\cos }^{2}}\theta }{\sin \theta \cos \theta {{\left( \sin \theta -\cos \theta \right)}^{2}}}$$ Now, it can be written as, $$\dfrac{\left( 1-2\sin \theta \cos \theta \right)\left( 1+\sin \theta \cos \theta \right)}{\sin \theta \cos \theta {{\left( \sin \theta -\cos \theta \right)}^{2}}}$$ Now, we will expand and write, $${{\left( \sin \theta -\cos \theta \right)}^{2}}\ =\ 1-2\sin \theta \cos \theta $$ So, we can write as, $$\dfrac{\left( 1-2\sin \theta \cos \theta \right)\left( 1+\sin \theta \cos \theta \right)}{\sin \theta \cos \theta \left( 1-2\sin \theta \cos \theta \right)}$$ Hence, the fraction can be written as, $$\dfrac{1+\sin \theta \cos \theta }{\sin \theta \cos \theta }$$ So, we will break into two fractions we get, $$\dfrac{1}{\sin \theta \cos \theta }+\dfrac{\sin \theta \cos \theta }{\sin \theta \cos \theta }$$ $$=\ \dfrac{1}{\sin \theta \cos \theta }+1=\ 1+\sec \theta \csc \theta $$ Hence, $$\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }$$is equal to $$1+\sec \theta \csc \theta $$ hence proved. Note: Student should know identities such as $${{\sin }^{2}}\theta +{{\cos }^{2}}\theta \ =\ 1$$, formula such as $$\tan \theta \ =\ \dfrac{\sin \theta }{\cos \theta }$$and$$\cot \theta \ =\ \dfrac{\cos \theta }{\sin \theta }$$.Making some necessary adjustments during simplification is key here.