Question

(1+ tan 33) (1 + tan (12)
(1 + tan 18') (1 + tans 27)​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{(1+tan\,33^\circ)\,(1+tan\,12^\circ)\,(1+tan\,18^\circ)\,(1+tan\,27^\circ)}$

$\underline{\textbf{To find:}}$

$\textsf{The value of}$

$\mathsf{(1+tan\,33^\circ)\,(1+tan\,12^\circ)\,(1+tan\,18^\circ)\,(1+tan\,27^\circ)}$

$\underline{\textbf{Solution:}}$

$\mathsf{Suppose,\;A+B=45^\circ}$

$\implies\mathsf{tan(A+B)=tan\,45^\circ}$

$\implies\mathsf{\dfrac{tan\,A+tan\,B}{1-tan\,A\;tan\,B}=1}$

$\implies\mathsf{tan\,A+tan\,B=1-tan\,A\;tan\,B}$

$\implies\mathsf{tan\,A+tan\,B+tan\,A\;tan\,B=1}$

$\implies\mathsf{1+tan\,A+tan\,B+tan\,A\;tan\,B=2}$

$\implies\mathsf{1(1+tan\,A)+tan\,B(1+tan\,A)=2}$

$\implies\mathsf{(1+tan\,A)\,(1+tan\,B)=2}$

$\mathsf{Thus,\;we\;have}$

$\boxed{\mathsf{If\;A+B=45^\circ,\;then\;\;(1+tan\,A)\,(1+tan\,B)=2}}$

$\mathsf{Now,}$

$\mathsf{(1+tan\,33^\circ)\,(1+tan\,12^\circ)\,(1+tan\,18^\circ)\,(1+tan\,27^\circ)}$

$\textsf{By the above result,}$

$\mathsf{(1+tan\,33^\circ)\,(1+tan\,12^\circ){\times}\,(1+tan\,18^\circ)\,(1+tan\,27^\circ)=2{\times}2}$

$\implies\boxed{\mathsf{(1+tan\,33^\circ)\,(1+tan\,12^\circ){\times}(1+tan\,18^\circ)\,(1+tan\,27^\circ)=4}}$

Answer 2

Answer:

We can solve this question using trigonometric formula and trigonometeric table. Final Answer: $\frac{(1+tan 33\textdegree) (1 + tan 12\textdegree)}{(1 + tan 18\textdegree) (1 + tan 27\textdegree)}=1$

Step-by-step explanation:

Correct question $\frac{(1+tan 33\textdegree) (1 + tan 12\textdegree)}{(1 + tan 18\textdegree) (1 + tan 27\textdegree)}$ =?

Step 1:

    $=\frac{(1+tan 33\textdegree) (1 + tan 12\textdegree)}{(1 + tan 18\textdegree) (1 + tan27\textdegree)}$

Step 2:

• $12\textdegree$ can be written as $45\textdegree-33\textdegree$ in the numerator and $27\textdegree$  can be written as  $45\textdegree-18\textdegree$ in the denominator .So that,

    $=\frac{(1+tan 33\textdegree) [1 + tan (45\textdegree-33\textdegree)]}{(1 + tan 18\textdegree) [1 + tan (45\textdegree-18\textdegree)]}$

Step 3:

• Formula: $tan(A-B)=\frac{tan A-tanB}{1+tanA.tanB}$
• $tan(45\textdegree-33\textdegree)=\frac{tan 45\textdegree-tan33\textdegree}{1+tan45\textdegree.tan33\textdegree}$
• $tan(45\textdegree-18\textdegree)=\frac{tan 45\textdegree-tan18\textdegree}{1+tan45\textdegree.tan18\textdegree}$

• Now apply the above formula in step 2.

    $=\frac{(1+tan 33\textdegree) [1 + (\frac{tan 45\textdegree-tan33\textdegree}{1+tan45\textdegree.tan33\textdegree})]}{(1 + tan 18\textdegree) [1 +( \frac{tan 45\textdegree-tan18\textdegree}{1+tan45\textdegree.tan18\textdegree})]}$

Step 4:  

• Value of $tan45\textdegree=1$ from trigonometeric table.

    $=\frac{(1+tan 33\textdegree) [1 + (\frac{1-tan33\textdegree}{1+1.tan33\textdegree})]}{(1 + tan 18\textdegree) [1 +( \frac{1-tan18\textdegree}{1+1.tan18\textdegree})]}$

    $=\frac{(1+tan 33\textdegree) [1 + (\frac{1-tan33\textdegree}{1+tan33\textdegree})]}{(1 + tan 18\textdegree) [1 +( \frac{1-tan18\textdegree}{1+tan18\textdegree})]}$

Step 5:

• Take the L.C.M for 1 and $tan33\textdegree$.

    $=\frac{(1+tan 33\textdegree) (\frac{1+tan33\textdegree+1-tan33\textdegree}{1+tan33\textdegree})}{(1 + tan 18\textdegree) ( \frac{1+tan18\textdegree+1-tan18\textdegree}{1+tan18\textdegree})}$

Step 6:

•  Add $tan 33\textdegree$ and $-tan33\textdegree$.Addition value is zero.

      $=\frac{(1+tan 33\textdegree) (\frac{1+1}{1+tan33\textdegree})}{(1 + tan 18\textdegree) ( \frac{1+1}{1+tan18\textdegree})}$

      $=\frac{(1+tan 33\textdegree) (\frac{2}{1+tan33\textdegree})}{(1 + tan 18\textdegree) ( \frac{2}{1+tan18\textdegree})}$

Step 7:

• Cancel $1+tan33\textdegree$ in the numerator and denominator.

      $=\frac{2}{2}$

      $=1$

Final Answer:  $\frac{(1+tan 33\textdegree) (1 + tan 12\textdegree)}{(1 + tan 18\textdegree) (1 + tan 27\textdegree)}=1$