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Question: Prove that \[\sin \theta (1 + \tan \theta ) + \cos \theta (1 + \cot \theta ) = \sec \theta + \csc \t...

Prove that sin⁡θ(1+tan⁡θ)+cos⁡θ(1+cot⁡θ)=sec⁡θ+csc⁡θ\sin \theta (1 + \tan \theta ) + \cos \theta (1 + \cot \theta ) = \sec \theta + \csc \theta .

Explanation

Solution

Hint : In this type of question we prove this by taking the left hand side of the equation and then solving this we get the right hand side of the equations. We can also take the right hand side of the equation and show it as the left hand side of the equation. We can expand the left hand side and can solve it easily.

Complete step-by-step answer :
Now, take LHS=sin⁡θ(1+tan⁡θ)+cos⁡θ(1+cot⁡θ)LHS = \sin \theta (1 + \tan \theta ) + \cos \theta (1 + \cot \theta ) and RHS=sec⁡θ+csc⁡θRHS = \sec \theta + \csc \theta
We know the identity in trigonometry,
tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} and
cot⁡θ=1tan⁡θ=cos⁡θsin⁡θ\cot \theta = \dfrac{1}{{\tan \theta }} = \dfrac{{\cos \theta }}{{\sin \theta }}
Substituting in LHSLHS, we get
⇒LHS=sin⁡θ(1+sin⁡θcos⁡θ)+cos⁡θ(1+cos⁡θsin⁡θ)\Rightarrow LHS = \sin \theta \left( {1 + \dfrac{{\sin \theta }}{{\cos \theta }}} \right) + \cos \theta \left( {1 + \dfrac{{\cos \theta }}{{\sin \theta }}} \right)
Expanding the bracket to simplify,
⇒LHS=sin⁡θ+sin⁡θ×sin⁡θcos⁡θ+cos⁡θ+cos⁡θ×cos⁡θsin⁡θ\Rightarrow LHS = \sin \theta + \sin \theta\times \dfrac{{\sin \theta }}{{\cos \theta }} + \cos \theta + \cos \theta\times \dfrac{{\cos \theta }}{{\sin \theta }}
Using simple multiplication, we get
⇒LHS=sin⁡θ+sin⁡2θcos⁡θ+cos⁡θ+cos⁡2θsin⁡θ\Rightarrow LHS = \sin \theta + \dfrac{{{{\sin }^2}\theta }}{{\cos \theta }} + \cos \theta + \dfrac{{{{\cos }^2}\theta }}{{\sin \theta }}
Rearranging the terms slightly we get,
⇒LHS=(sin⁡θ+cos⁡2θsin⁡θ)+(cos⁡θ+sin⁡2θcos⁡θ)\Rightarrow LHS = \left( {\sin \theta + \dfrac{{{{\cos }^2}\theta }}{{\sin \theta }}} \right) + \left( {\cos \theta + \dfrac{{{{\sin }^2}\theta }}{{\cos \theta }}} \right)
We rearranged the above equation such a way that we will get a simplified answer and we can see that after taking L.C.M we can use a trigonometric identity. If we rearranged above in other forms it will be difficult to solve and takes lots of time and steps.
That is,
Further taking L.C.M and simplifying we get,
⇒LHS=(sin⁡2θ+cos⁡2θsin⁡θ)+(cos⁡2θ+sin⁡2θcos⁡θ)\Rightarrow LHS = \left( {\dfrac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{\sin \theta }}} \right) + \left( {\dfrac{{{{\cos }^2}\theta + {{\sin }^2}\theta }}{{\cos \theta }}} \right)
We know that, the basic identity in trigonometry,
sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
Substituting these in above,
Above becomes, ⇒LHS=(1sin⁡θ)+(1cos⁡θ) \Rightarrow LHS = \left( {\dfrac{1}{{\sin \theta }}} \right) + \left( {\dfrac{1}{{\cos \theta }}} \right)
We know that, 1sin⁡θ=csc⁡θ\dfrac{1}{{\sin \theta }} = \csc \theta and 1cos⁡θ=sec⁡θ\dfrac{1}{{\cos \theta }} = \sec \theta that is reciprocals,
⇒LHS=csc⁡θ+sec⁡θ\Rightarrow LHS = \csc \theta + \sec \theta
That is⇒LHS=RHS \Rightarrow LHS = RHS,
Thus we show that sin⁡θ(1+tan⁡θ)+cos⁡θ(1+cot⁡θ)=sec⁡θ+csc⁡θ\sin \theta (1 + \tan \theta ) + \cos \theta (1 + \cot \theta ) = \sec \theta + \csc \theta .
Hence proved.

Note : In mathematics, trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables where both sides of the equality are defined. Geometrically, these are identities involving certain functions of one or more angles.