If (1+tan 1^{circ})(1+tan 2^{circ}) ldots (1+ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Consider $A+B=45^{\circ}$. Then $	an(A+B)=1$,which implies $	an A + 	an B = 1 - 	an A 	an B$. Adding $1 + 	an A 	an B$ to both sides gives

Vedclass · Accepted Answer

$23$

Vedclass · Answer

(C) Consider $A+B=45^{\circ}$. Then $	an(A+B)=1$,which implies $	an A + 	an B = 1 - 	an A 	an B$. Adding $1 + 	an A 	an B$ to both sides gives $1 + 	an A + 	an B + 	an A 	an B = 2$,or $(1+	an A)(1+	an B) = 2$. We can pair terms from $1^{\circ}$ to $44^{\circ}$ such that the sum of angles is $45^{\circ}$: $(1+	an 1^{\circ})(1+	an 44^{\circ}) = 2$,$(1+	an 2^{\circ})(1+	an 43^{\circ}) = 2$,...,$(1+	an 22^{\circ})(1+	an 23^{\circ}) = 2$. There are $22$ such pairs,so their product is $2^{22}$. Finally,we include the last term: $(1+	an 45^{\circ}) = 1+1 = 2$. Thus,the total product is $2^{22} 	imes 2 = 2^{23}$. Comparing with $2^n$,we get $n=23$.

If $(1+\tan 1^{\circ})(1+\tan 2^{\circ}) \ldots (1+\tan 45^{\circ})=2^n$,then $n=$

Similar Questions

$\cot 18^{\circ} \cdot \cot 36^{\circ}+1=$

If $\cos \alpha + \cos \beta + \cos \gamma = 0$ and $\sin \alpha + \sin \beta + \sin \gamma = 0$,then $\cos 2\alpha + \cos 2\beta + \cos 2\gamma$ equals:

$\sinh (x+y) \cosh (x-y)$ is equal to

In triangles $ABC$ and $DEF$,$AB = DE$,$AC = EF$ and $\angle A = 2\angle E$. The two triangles will have the same area if angle $A$ is equal to:

If $\tan \alpha = \frac{x^2 - x}{x^2 - x + 1}$ and $\tan \beta = \frac{1}{2x^2 - 2x + 1}$ $(x \ne 0, 1)$,where $0 < \alpha, \beta < \frac{\pi}{2}$,then $\tan(\alpha + \beta)$ has the value equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module