# Solve the trigonometrical product

solve this equation :

$(2-\sec^21)(2-\sec^22)(2-\sec^23)........(2-\sec^288)(2-\sec^289)$

If tried from tangent approach with$(1+1-sec^21)......(1+1-\sec^289)$

and i do (1,89) ; (2,88);and........ so on so i get

$(1-\tan^21)(1-\tan^289)$

$(1+\tan^21\tan^289-\tan^289-\tan^21)$

i've got $\tan^21\tan^289= 1$

from $\tan(89+1)=\frac{\tan89+\tan1}{1-\tan89\tan1}$

but i don't know how to get $(\tan^289+\tan^21)$

and i don't know how to continue it

and please from other approach too

thank you

You get $$\prod_{0\le x\le 89}(1-\tan^2x^\circ)\text{ not }\prod_{0\le x\le 89}(1+\tan^2x^\circ)$$
But $\tan45^\circ=1$
Hint: What is $\cos 45^{\circ}$?
• $\frac{2^{\frac{1}{2}}}{2}$ but what is the relationship? Apr 17, 2013 at 8:18
• @freeze: I suggest you don't just stop there, and see how you might use that hint. Here is another puzzle of the same kind: what is $(x-a)(x-b)(x-c)\dots(x-z)?$ Apr 17, 2013 at 8:21
• but i don't know how to get $(\tan^289+tan^21)$? can you help me Apr 17, 2013 at 8:24
• @freeze: What's the value of this : $(4-1)(4-2)(4-3)(4-4)(4-5)..(4-n)$? Apr 17, 2013 at 9:34