যৌগিক কোন সম্বলিত ত্রিকোণমিতিক রাশি
cotα+cotβ=a,tanα+tanβ=b \cot \alpha+\cot \beta=a, \tan \alpha+\tan \beta=b cotα+cotβ=a,tanα+tanβ=b এবং α+β=θ \alpha+\beta=\theta α+β=θহলে নিচের কোনটি সঠিক?
(a−b)tanθ=ab (a-b) \tan \theta=a b(a−b)tanθ=ab
(a+b)tanθ=ab (a+b) \tan \theta=a b(a+b)tanθ=ab
(ab)tanθ=a+b (ab) \tan \theta=a+ b(ab)tanθ=a+b
(a−b)tanθ=−ab (a-b) \tan \theta=-a b(a−b)tanθ=−ab
Solve: দেওয়া আছে,cotα+cotβ=a⋯(1),tanα+tanβ=b⋯(2) এবং α+β=θ⋯ \begin{array}{l} \cot \alpha+\cot \beta=a \cdots(1), \\ \tan \alpha+\tan \beta=b \cdots(2) \text { এবং } \alpha+\beta=\theta \cdots \end{array} cotα+cotβ=a⋯(1),tanα+tanβ=b⋯(2) এবং α+β=θ⋯
(1) হতে. আমরা পাই , 1tanα+1tanβ=a \frac{1}{\tan \alpha}+\frac{1}{\tan \beta}=a tanα1+tanβ1=a
⇒tanβ+tanαtanαtanβ=a⇒btanαtanβ=a⇒tanαtanβ=ba এখন , θ=α+β⇒tanθ=tan(α+β)=tanα+tanβ1−tanαtanβ=b1−ba=aba−b∴(a−b)tanθ=ab \begin{array}{l} \Rightarrow \frac{\tan \beta+\tan \alpha}{\tan \alpha \tan \beta}=a \\ \Rightarrow \frac{b}{\tan \alpha \tan \beta}=a \Rightarrow \tan \alpha \tan \beta=\frac{b}{a} \\ \text { এখন , } \theta=\alpha+\beta \\ \Rightarrow \tan \theta=\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \\ =\frac{b}{1-\frac{b}{a}}=\frac{a b}{a-b} \\ \therefore \quad(a-b) \tan \theta=a b \\ \end{array} ⇒tanαtanβtanβ+tanα=a⇒tanαtanβb=a⇒tanαtanβ=ab এখন , θ=α+β⇒tanθ=tan(α+β)=1−tanαtanβtanα+tanβ=1−abb=a−bab∴(a−b)tanθ=ab
If sinα+sinβ=12sin \alpha + sin \beta = \dfrac{1}{2} sinα+sinβ=21 and cosα+cosβ=32cos \alpha + cos \beta = \dfrac{\sqrt{3}}{2}cosα+cosβ=23 then 3β+α=3 \beta + \alpha = 3β+α=
2tan(45∘+x)1+tan2(45∘+x)= \frac{2 \tan \left(45^{\circ}+x\right)}{1+\tan ^{2}\left(45^{\circ}+x\right)}= 1+tan2(45∘+x)2tan(45∘+x)= কত?
tanα−tanβ=x \tan \alpha-\tan \beta=x tanα−tanβ=x এবং cotβ−cotα=y \cot \beta-\cot \alpha=y cotβ−cotα=y হলে নিচের কোনটি সঠিক, cot(α−β)=? \cot (\alpha-\beta)= ? cot(α−β)=?
Given that
(1+(1+x))tanx=1+(1−x)(1+\sqrt{(1+x)})\tan x=1+\sqrt{(1-x)}(1+(1+x))tanx=1+(1−x)
If 4x4x4x is in second quadrant, then sin4x\sin 4xsin4x is equal to
