Question

If $\frac{1}{\sin {45}^{\circ }\sin {46}^{\circ }}+\frac{1}{\sin {47}^{\circ }\sin {48}^{\circ }}+...+\frac{1}{\sin {135}^{\circ }\sin {134}^{\circ }}=\frac{1}{\sin A^{\circ }}$ and
$\frac{1}{\sin 1^{\circ }\sin 2^{\circ }}+\frac{1}{\sin 2^{\circ }\sin 3^{\circ }}+...+\frac{1}{\sin {89}^{\circ }\sin {90}^{\circ }}=\frac{B}{\sin 1^{\circ }},$
then which of the following is/are correct ?

• A. The smallest positive integral value of $A$ is $1.$
• B. The smallest positive integral value of $A$ is $45.$
• C. $B=\cos 1^{\circ }$
• D. $B=\cot 1^{\circ }$

Answer 1

Answer: (A), (D)

The correct options are
A The smallest positive integral value of $A$ is $1.$
D $B=\cot 1^{\circ }$

$\sin 1^{\circ }=\sin \left[\right(x+1{)}^{\circ }-x^{\circ }]$
$=\sin (x+1{)}^{\circ }\cos x^{\circ }-\cos (x+1{)}^{\circ }\sin x^{\circ }$

$\frac{\sin 1^{\circ }}{\sin x^{\circ }\sin (x+1{)}^{\circ }}=\frac{\cos x^{\circ }\sin (x+1{)}^{\circ }-\sin x^{\circ }\cos (x+1{)}^{\circ }}{\sin x^{\circ }\sin (x+1{)}^{\circ }}$

$\Rightarrow \frac{\sin 1^{\circ }}{\sin x^{\circ }\sin (x+1{)}^{\circ }}=\cot x^{\circ }-\cot (x+1{)}^{\circ }$

$\frac{1}{\sin {45}^{\circ }\sin {46}^{\circ }}+\frac{1}{\sin {47}^{\circ }\sin {48}^{\circ }}+...+\frac{1}{\sin {135}^{\circ }\sin {134}^{\circ }}=\frac{1}{\sin A^{\circ }}$

Multiplying both sides by $\sin 1^{\circ },$ we get

$\frac{\sin 1^{\circ }}{\sin A^{\circ }}=(\cot {45}^{\circ }-\cot {46}^{\circ })+(\cot {47}^{\circ }-\cot {48}^{\circ })$
$+...+(\cot {133}^{\circ }-\cot {134}^{\circ })$
$=\cot {45}^{\circ }-(\cot {46}^{\circ }+\cot {134}^{\circ })+(\cot {47}^{\circ }+\cot {133}^{\circ })$
$-...+(\cot {89}^{\circ }+\cot {91}^{\circ })-\cot {90}^{\circ }$
$=1$
$\Rightarrow \sin A^{\circ }=\sin 1^{\circ }$
So, the least possible integer value for $A$ is $1.$


$\frac{1}{\sin 1^{\circ }\sin 2^{\circ }}+\frac{1}{\sin 2^{\circ }\sin 3^{\circ }}+...+\frac{1}{\sin {89}^{\circ }\sin {90}^{\circ }}=\frac{B}{\sin 1^{\circ }}$

$\text{LHS}=\sum _{k=1}^{89}\frac{1}{\sin k^{\circ }\sin (k+1{)}^{\circ }}$
$=\frac{1}{\sin 1^{\circ }}\sum _{k=1}^{89}[\cot k^{\circ }-\cot (k+1{)}^{\circ }]$
$=\frac{\cot 1^{\circ }}{\sin 1^{\circ }}$

$\Rightarrow B=\cot 1^{\circ }$