Solveeit Logo
Login

Question

Question: The value of \[\cot {15^ \circ }\cot {16^ \circ }\cot {17^ \circ }...\cot {73^ \circ }\cot {74^ \cir...

The value of cot⁡15∘cot⁡16∘cot⁡17∘...cot⁡73∘cot⁡74∘cot⁡75∘\cot {15^ \circ }\cot {16^ \circ }\cot {17^ \circ }...\cot {73^ \circ }\cot {74^ \circ }\cot {75^ \circ }
A.12\dfrac{1}{2}
B. 0
C. 1
D. -1

Explanation

Solution

Here we convert the angles of the first thirty terms such that their sum is equal to 90∘{90^ \circ }. We use the concept of complementary angles. Here cot and tan are complementary angles. Multiply the cot and tan of the same angles respectively to get the value.

  • Complementary angles means that tan⁡(90∘−θ)=cot⁡θ\tan ({90^ \circ } - \theta ) = \cot \theta andcot⁡(90∘−θ)=tan⁡θ\cot ({90^ \circ } - \theta ) = \tan \theta
  • tan⁡θ=1cot⁡θ\tan \theta = \dfrac{1}{{\cot \theta }}

Complete step-by-step answer:
We have the equation cot⁡15∘cot⁡16∘cot⁡17∘...cot⁡73∘cot⁡74∘cot⁡75∘\cot {15^ \circ }\cot {16^ \circ }\cot {17^ \circ }...\cot {73^ \circ }\cot {74^ \circ }\cot {75^ \circ }
Total number of values in the multiplication is75−15=6075 - 15 = 60
So we write the terms as
⇒cot⁡15∘cot⁡16∘cot⁡17∘...cot⁡44∘cot⁡45∘cot⁡46∘...cot⁡73∘cot⁡74∘cot⁡75∘\Rightarrow \cot {15^ \circ }\cot {16^ \circ }\cot {17^ \circ }...\cot {44^ \circ }\cot {45^ \circ }\cot {46^ \circ }...\cot {73^ \circ }\cot {74^ \circ }\cot {75^ \circ }
So we write the first 30 terms in terms of complimentary angles.
⇒cot⁡(90∘−15∘)cot⁡(90∘−16∘)cot⁡(90∘−17∘)...cot⁡(90∘−44∘)cot⁡(90∘−45∘)cot⁡46∘.....cot⁡73∘cot⁡74∘cot⁡75∘\Rightarrow \cot ({90^ \circ } - {15^ \circ })\cot ({90^ \circ } - {16^ \circ })\cot ({90^ \circ } - {17^ \circ })...\cot ({90^ \circ } - {44^ \circ })\cot ({90^ \circ } - {45^ \circ })\cot {46^ \circ }.....\cot {73^ \circ }\cot {74^ \circ }\cot {75^ \circ }
Use the concept of complimentary angles i.e. cot⁡(90∘−θ)=tan⁡θ\cot ({90^ \circ } - \theta ) = \tan \theta
Then we can write
cot⁡(90∘−15∘)=tan⁡75∘\cot ({90^ \circ } - {15^ \circ }) = \tan {75^ \circ },cot⁡(90∘−16∘)=tan⁡74∘\cot ({90^ \circ } - {16^ \circ }) = \tan {74^ \circ },cot⁡(90∘−17∘)=tan⁡73∘\cot ({90^ \circ } - {17^ \circ }) = \tan {73^ \circ }...cot⁡(90∘−44∘)=tan⁡46∘\cot ({90^ \circ } - {44^ \circ }) = \tan {46^ \circ },cot⁡(90∘−45∘)=tan⁡45∘\cot ({90^ \circ } - {45^ \circ }) = \tan {45^ \circ }
Then the equation becomes
⇒tan⁡75∘tan⁡74∘tan⁡73∘...tan⁡46∘tan⁡45∘cot⁡46∘...cot⁡73∘cot⁡74∘cot⁡75∘\Rightarrow \tan {75^ \circ }\tan {74^ \circ }\tan {73^ \circ }...\tan {46^ \circ }\tan {45^ \circ }\cot {46^ \circ }...\cot {73^ \circ }\cot {74^ \circ }\cot {75^ \circ }
Substitute the value of tan⁡45∘=1\tan {45^ \circ } = 1
⇒tan⁡75∘tan⁡74∘tan⁡73∘...tan⁡46∘cot⁡46∘...cot⁡73∘cot⁡74∘cot⁡75∘\Rightarrow \tan {75^ \circ }\tan {74^ \circ }\tan {73^ \circ }...\tan {46^ \circ }\cot {46^ \circ }...\cot {73^ \circ }\cot {74^ \circ }\cot {75^ \circ }
Now we pair the terms having same angles
⇒(tan⁡75∘cot⁡75∘)(tan⁡74∘cot⁡74∘)(tan⁡73∘cot⁡73∘)...(tan⁡46∘cot⁡46∘)\Rightarrow (\tan {75^ \circ }\cot {75^ \circ })(\tan {74^ \circ }\cot {74^ \circ })(\tan {73^ \circ }\cot {73^ \circ })...(\tan {46^ \circ }\cot {46^ \circ })
Use the formula tan⁡θ=1cot⁡θ\tan \theta = \dfrac{1}{{\cot \theta }} and change every value in tan to cot in the equation.
⇒(1cot⁡75∘cot⁡75∘)(1cot⁡74∘cot⁡74∘)(1cot⁡73∘cot⁡73∘)...(1cot⁡46∘cot⁡46∘)\Rightarrow (\dfrac{1}{{\cot {{75}^ \circ }}}\cot {75^ \circ })(\dfrac{1}{{\cot {{74}^ \circ }}}\cot {74^ \circ })(\dfrac{1}{{\cot {{73}^ \circ }}}\cot {73^ \circ })...(\dfrac{1}{{\cot {{46}^ \circ }}}\cot {46^ \circ })
Cancel out same terms from numerator and denominator from each pair
⇒(1)(1)(1)...(1)=1\Rightarrow (1)(1)(1)...(1) = 1
Therefore, the value of the given equation is 1

So, option C is correct.

Note: Students might get confused as of how many angles are we changing in complimentary terms. Keep in mind we calculate the total number of terms in multiplication and divide them in equal halves, so every value in the left half will have a complimentary value in the right half of the equation. Also, the middle most term with angle 45∘{45^ \circ }can be written directly with the knowledge of trigonometric terms at major angles. For those who don’t remember the values, they can take help from the following table.

Angles (in degrees)0∘{0^ \circ }30∘{30^ \circ }45∘{45^ \circ }60∘{60^ \circ }90∘{90^ \circ }
sin012\dfrac{1}{2}12\dfrac{1}{{\sqrt 2 }}32\dfrac{{\sqrt 3 }}{2}11
cos132\dfrac{{\sqrt 3 }}{2}12\dfrac{1}{{\sqrt 2 }}12\dfrac{1}{2}0
tan013\dfrac{1}{{\sqrt 3 }}13\sqrt 3 Not defined