মান নির্ণয়
যদি cos x+cos y=a এবং sin x + sin y = b হয় তবে cos (x + y) এর মান কোনটি?
a2+b2a−b \frac{a ² + b ²}{a - b} a−ba2+b2
a2−b2a2+b2 \frac{a ² - b ²}{a ² + b ²} a2+b2a2−b2
a2=cos2x+cos2y+2cosx⋅cosy;b2=sin2x+sin2y+2sinx⋅siny∴a2−b2=cos2x+cos2y+2cos(x+y);a2+b2=2+2cos(x−y) এখন, a2−b2=2cos(x+y)⋅cos(x−y)+2cos(x+y)=cos(x+y){2cos(x−y)+2}=cos(x+y)×(a2+b2)∴cos(x+y)=a2−b2a2+b2 \begin{array}{l}\text { } a^{2}=\cos ^{2} x+\cos ^{2} y+2 \cos x \cdot \cos y ; b^{2}=\sin ^{2} x+\sin ^{2} y+2 \sin x \cdot \sin y \\ \therefore a^{2}-b^{2}=\cos 2 x+\cos 2 y+2 \cos (x+y) ; a^{2}+b^{2}=2+2 \cos (x-y) \\ \text { এখন, } a^{2}-b^{2}=2 \cos (x+y) \cdot \cos (x-y)+2 \cos (x+y) \\ =\cos (x+y)\{2 \cos (x-y)+2\}=\cos (x+y) \times\left(a^{2}+b^{2}\right) \therefore \cos (x+y)=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\end{array} a2=cos2x+cos2y+2cosx⋅cosy;b2=sin2x+sin2y+2sinx⋅siny∴a2−b2=cos2x+cos2y+2cos(x+y);a2+b2=2+2cos(x−y) এখন, a2−b2=2cos(x+y)⋅cos(x−y)+2cos(x+y)=cos(x+y){2cos(x−y)+2}=cos(x+y)×(a2+b2)∴cos(x+y)=a2+b2a2−b2
tanθ=p হলে, cos2θ= কত? \tan \theta=p \text { হলে, } \cos 2 \theta=\text { কত? } tanθ=p হলে, cos2θ= কত?
যদি π2<θ<πএবংsinθ=35হয়, \frac{\pi}{2} < \theta < \pi এ ব ং \sin{\theta} = \frac{3}{5} হ য় , 2π<θ<πএবংsinθ=53হয়, তবে cosθ এর মান কত?
If cosθ=513\displaystyle \cos \theta =\frac{5}{13}cosθ=135, where θ\theta θ being an acute angle, then the value of cosθ+5cotθcosec θ−cosθ\dfrac{\cos \theta +5\cot \theta }{\text {cosec}\ \theta -\cos \theta }cosec θ−cosθcosθ+5cotθ will be
What is cot(A2)−tan(A2)\cot(\dfrac{A}{2})-\tan(\dfrac{A}{2})cot(2A)−tan(2A) equal to ?
