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Question: \(2n\pi + \frac{5\pi}{4}\), n is integer), if \((2n + 1)\pi + \frac{\pi}{4}\)...

2nπ+5π42n\pi + \frac{5\pi}{4}, n is integer), if (2n+1)π+π4(2n + 1)\pi + \frac{\pi}{4}

A

sin(A+B)=1\sin(A + B) = 1 and cos(AB)=32\cos(A - B) = \frac{\sqrt{3}}{2}

B

\Rightarrowand A+B=π2A + B = \frac{\pi}{2}

C

AB=π6A - B = \frac{\pi}{6}only

D

\Rightarrowonly

Answer

\Rightarrowand A+B=π2A + B = \frac{\pi}{2}

Explanation

Solution

f(x)f(x)

2n2(n1)2n2(n - 1) or 2n(n1)\Rightarrow 2n(n - 1)

\Rightarrow tanC2=(sa)(sb)s(sc)=1\tan\frac{C}{2} = \sqrt{\frac{(s - a)(s - b)}{s(s - c)}} = 1 or =tan(π4)= \tan\left( \frac{\pi}{4} \right)

\Rightarrow sinA2=(sb)(sc)bc\sin\frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{bc}} and bcsin2A2=(sb)(sc)\Rightarrow bc\sin^{2}\frac{A}{2} = (s - b)(s - c).