Question

Solve the trigonometric expression:
\[\dfrac{\cos 35{}^\circ }{\sin 55{}^\circ }+\dfrac{\tan 27{}^\circ \tan 63{}^\circ }{\sin 30{}^\circ }-3{{\left( \tan 60{}^\circ \right)}^{2}}\].

Answer 1

Hint: Now for solving this we must know the complementary functions of trigonometry and the formulas of all the functions too so that we can easily solve and simplify this question. In the end we also need to know the trigonometric values of basic angles for all functions when the angles are like \[0{}^\circ ,30{}^\circ ,45{}^\circ ,60{}^\circ ,90{}^\circ ,180{}^\circ \].

Complete step-by-step solution:
Now starting the solution. The expression given to us here in this question is:
\[\dfrac{\cos 35{}^\circ }{\sin 55{}^\circ }+\dfrac{\tan 27{}^\circ \tan 63{}^\circ }{\sin 30{}^\circ }-3{{\left( \tan 60{}^\circ \right)}^{2}}\]
To solve this question we will start its simplification step by step from the first term to the last term so as to make it easier to understand
Now we know that cosine is a complementary angle to sine which means that \[\cos (90-x)=\sin x\], therefore with that formula and knowledge we can find the complement of cos in first term
\[\dfrac{\sin (90{}^\circ -35{}^\circ )}{\sin 55{}^\circ }+\dfrac{\tan 27{}^\circ \tan 63{}^\circ }{\sin 30{}^\circ }-3{{\left( \tan 60{}^\circ \right)}^{2}}\]
Solving
\[\dfrac{\sin 55{}^\circ }{\sin 55{}^\circ }+\dfrac{\tan 27{}^\circ \tan 63{}^\circ }{\sin 30{}^\circ }-3{{\left( \tan 60{}^\circ \right)}^{2}}\]
Since both numerator and denominator are the same we can cancel them and get
\[1+\dfrac{\tan 27{}^\circ \tan 63{}^\circ }{\sin 30{}^\circ }-3{{\left( \tan 60{}^\circ \right)}^{2}}\]
Now therefore taking the second term. Now here in the denominator we can see that we are asked the value of \[\sin 30{}^\circ \] which is a basic angle so putting its value we are left with
\[1+\dfrac{\tan 27{}^\circ \tan 63{}^\circ }{\dfrac{1}{2}}-3{{\left( \tan 60{}^\circ \right)}^{2}}\]
Since we know that cot is the complementary angle of tan which means that \[\tan (90-x)=\cot x\] therefore using that
\[1+2\left( \tan 27{}^\circ \tan (90{}^\circ -27{}^\circ ) \right)-3{{\left( \tan 60{}^\circ \right)}^{2}}\]
Now solving further
\[1+2\left( \tan 27{}^\circ \cot 27{}^\circ \right)-3{{\left( \tan 60{}^\circ \right)}^{2}}\]
Through trigonometry formulas we know that \[\cot x=\dfrac{1}{\tan x}\] therefore using that formula here
\[1+2\left( \tan 27{}^\circ \times \dfrac{1}{\tan 27{}^\circ } \right)-3{{\left( \tan 60{}^\circ \right)}^{2}}\]
Now cancelling the same terms
\[1+2-3{{\left( \tan 60{}^\circ \right)}^{2}}\]
Again in last term we can directly put the value of \[\tan 60{}^\circ \] that we know and take the square and solve it
\[1+2-3{{\left( \sqrt{3} \right)}^{2}}\]
Taking square and multiplying by opening the bracket
\[=1+2-9\]
Solving this
\[=-6\]
Therefore the answer for the question is \[\dfrac{\cos 35{}^\circ }{\sin 55{}^\circ }+\dfrac{\tan 27{}^\circ \tan 63{}^\circ }{\sin 30{}^\circ }-3{{\left( \tan 60{}^\circ \right)}^{2}}=-6\].

Note: Now here in this question we need to always beware of writing the formula for complementary angles or else some simple mistake could make the students lost marks. Some common mistakes that might be made are
\[\tan (90-27)=\cot 23\] ;this is the correct way to write the formula
\[\tan (90-63)\ne \cot 23\]; this is the wrong way to write the formula