TITLE: How to find $(2-\sec^2 1^\circ)(2-\sec^2 2^\circ)\cdots \overline{(2-\sec^2 45^\circ)}\cdots(2-\sec^2 89^\circ)$
QUESTION [5 upvotes]: Evaluate
$$(2-\sec^2{1^{\circ}})(2-\sec^2{2^{\circ}})(2-\sec^2{3^{\circ}})\cdots(2-\sec^2{44^{\circ}})(2-\sec^2{46^{\circ}})\cdots(2-\sec^2{89^{\circ}})$$
This same problem come from Problem 21:But that problem is very very easy http://purplecomet.org/home/resource/544/HighSchoolSolutions2013.pdf
$$(2-\sec^2{1^{\circ}})(2-\sec^2{2^{\circ}})(2-\sec^2{3^{\circ}})\cdots\cdots(2-\sec^2{89^{\circ}})$$
and since $sec^2{45^{\circ}}=2$
so
$$(2-\sec^2{1^{\circ}})(2-\sec^2{2^{\circ}})(2-\sec^2{3^{\circ}})\cdots(2-\sec^2{89^{\circ}})=0$$
so I ask this follow problem have value?:$$(2-\sec^2{1^{\circ}})(2-\sec^2{2^{\circ}})(2-\sec^2{3^{\circ}})\cdots(2-\sec^2{44^{\circ}})(2-\sec^2{46^{\circ}})\cdots(2-\sec^2{89^{\circ}})$$
Thank you, and this problem is interesting.
we kown that $$2-\sec^2{x}=\dfrac{\sec^2{x}}{\sec{2x}}$$
and we kown 
$$\sin{1^{0}}\sin{2^{0}}\cdots\sin{89^{0}}=\dfrac{6}{4^{45}}\sqrt{10}$$
This  result
use this $$\sin{3x}=4\sin{x}\sin{(x+60^{0})}\sin{(60^{0}-x)}$$
so
$$\cos{1^{0}}\cos{2^{0}}\cdots\cos{89^{0}}=\dfrac{6}{4^{45}}\sqrt{10}$$
so
\begin{align}
&(2-\sec^2{1^{\circ}})(2-\sec^2{2^{\circ}})(2-\sec^2{3^{\circ}})\cdots(2-\sec^2{44^{\circ}})(2-\sec^2{46^{\circ}})\cdots(2-\sec^2{89^{0}})\\
&=\dfrac{cos{2^{0}}\cos{4^{0}}\cdots\cos{88^{0}}(-\sin{2^{0}})(-\sin{4^{0}})\cdots(-\sin{88^{0}})}{\cos^2{1^{0}}\cos^2{2^{0}}\cdots\cos^2{89^{0}}}\\
&=\dfrac{(\sin{4^{0}}\sin{8^{0}}\cdots\sin{88^{0}})^2}{A^2}
\end{align}
where $A=\dfrac{12}{4^{45}}\sqrt{5}$
so we must find $\sin{4^{0}}\sin{8^{0}}\cdots\sin{88^{0}}$
I think must use
$$\sin{3x}=4\sin{x}\sin{(60-x)}\sin{(60+x)}$$
Thank you, I have solve it.
I have this result 
$$(2-\sec^2{1^{\circ}})(2-\sec^2{2^{\circ}})(2-\sec^2{3^{\circ}})\cdots(2-\sec^2{44^{\circ}})(2-\sec^2{46^{\circ}})\cdots(2-\sec^2{89^{\circ}})=2^{88}$$

REPLY [3 votes]: Actual problem is to find the value of the Product $P$ and $Q$ where 
$$ P= \left(1-{Tan^2{1^\circ}}\right) \left(1-{Tan^2{2^\circ}}\right)\cdots \left(1-{Tan^2{44^\circ}}\right) $$ 
$$ Q= \left(1-{Tan^2{46^\circ}}\right) \left(1-{Tan^2{47^\circ}}\right)\cdots \left(1-{Tan^2{89^\circ}}\right) $$ $\implies$
$$ Q= \left(1-{Cot^2{44^\circ}}\right) \left(1-{Cot^2{43^\circ}}\right)\cdots \left(1-{Cot^2{1^\circ}}\right) $$ 
Let $$ R= Tan{1^\circ} \:Tan{2^\circ}\cdots Tan{44^\circ} $$ So
$$ Q=\frac{P}{R^2}$$ Also
$$ S=\left(1+Tan{1^\circ}\right)\left(1+Tan{2^\circ}\right)\cdots \left(1+Tan{44^\circ}\right)=2^{22} $$ and
$$ T=\left(1-Tan{1^\circ}\right)\left(1-Tan{2^\circ}\right)\cdots \left(1-Tan{44^\circ}\right) $$ so
$$ T=\left(\frac{2Tan{44^\circ}}{1+Tan{44^\circ}}\right) \left(\frac{2Tan{43^\circ}}{1+Tan{43^\circ}}\right) \cdots \left(\frac{2Tan{1^\circ}}{1+Tan{1^\circ}}\right)$$ So
$$ T=\frac{2^{44}R}{S}=2^{22}R$$ But
$$P=ST=2^{44}R$$ So
$$Q=\frac{P}{R^2}=\frac{2^{44}}{R}$$ Finally
$$PQ=2^{88}$$