মান নির্ণয়
sinA=12,sinB=13\sin A=\frac{1}{\sqrt{2}},\sin B=\frac{1}{\sqrt{3}}sinA=21,sinB=31হলে tan(A+B)=\tan(A+B)=tan(A+B)=কত?
1−21+2\frac{1-\sqrt{2}}{1+\sqrt{2}}1+21−2
1+21−2\frac{1+\sqrt{2}}{1-\sqrt{2}}1−21+2
2−12+1\frac{\sqrt{2}-1}{\sqrt{2}+1}2+12−1
2+12−1\frac{\sqrt{2}+1}{\sqrt{2}-1}2−12+1
∴cosA=12 \therefore \operatorname{cos} A=\frac{1}{\sqrt{2}} ∴cosA=21
cosB=23 \cos B=\frac{\sqrt{2}}{\sqrt{3}} cosB=32
∴tan(A+B)=tanA+tanB1−tanAtanB \therefore \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B} ∴tan(A+B)=1−tanAtanBtanA+tanB
=sinAcosA+sinBcosB1−sinAcosA⋅sinBcosB =\frac{\frac{\sin A}{\operatorname{cos} A}+\frac{\sin B}{\cos B}}{1-\frac{\sin A}{\cos A} \cdot \frac{\sin B}{\cos B}} =1−cosAsinA⋅cosBsinBcosAsinA+cosBsinB
=1+121−1⋅12=2+12−1 \begin{array}{l}=\frac{1+\frac{1}{\sqrt{2}}}{1-1 \cdot \frac{1}{\sqrt{2}}} \\ =\frac{\sqrt{2}+1}{\sqrt{2}-1}\end{array} =1−1⋅211+21=2−12+1
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?
tan105∘=tan(60∘+45∘)\tan 105^{\circ}=\tan \left(60^{\circ}+45^{\circ}\right)tan105∘=tan(60∘+45∘) 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
