ত্রিকোণমিতিক কোণের মধ্যে সম্পর্ক
acosθ+bsinθ=ca\cos{\theta}+b\sin{\theta}=cacosθ+bsinθ=c সমীকরণের বাস্তব সমাধান থাকলে c=?c=?c=?
c=a2+b2c=\sqrt{a^2+b^2} c=a2+b2
c>a2+b2c>\sqrt{a^2+b^2} c>a2+b2
c≤(a2+b2)c\le\left(a^2+b^2\right) c≤(a2+b2)
∣c∣≤a2+b2|c|\le\sqrt{a^2+b^2} ∣c∣≤a2+b2
a cosθ +b sinθ = ca\ \cos\theta\ +b\ \sin\theta\ =\ ca cosθ +b sinθ = c
⇒a cosθ= c −b sinθ\Rightarrow a\ \cos\theta=\ c\ -b\ \sin\theta⇒a cosθ= c −b sinθ
⇒a2 cos2θ = c2 −2bc sinθ +b2 sin2θ\Rightarrow a^2\ \cos^2\theta\ =\ c^{2\ }-2bc\ \sin\theta\ +b^2\ \sin^2\theta⇒a2 cos2θ = c2 −2bc sinθ +b2 sin2θ
⇒⇒a2 (1−sin2θ )= c2 −2bc sinθ +b2 sin2θ\Rightarrow\Rightarrow a^2\ \left(1-\sin^2\theta\ \right)=\ c^{2\ }-2bc\ \sin\theta\ +b^2\ \sin^2\theta⇒⇒a2 (1−sin2θ )= c2 −2bc sinθ +b2 sin2θ
⇒(a2+b2)sin2θ − 2bc sinθ +(c2 −a2 )\Rightarrow\left(a^2+b^2\right)\sin^2\theta\ -\ 2bc\ \sin\theta\ +\left(c^2\ -a^2\ \right)⇒(a2+b2)sin2θ − 2bc sinθ +(c2 −a2 )
বাস্তব মান থাকলে, (2bc)2−4(a2+b2).(c2−a2)≥0\left(2bc\right)^2-4\left(a^2+b^2\right).\left(c^2-a^2\right)\ge0(2bc)2−4(a2+b2).(c2−a2)≥0
⇒b2c2≥qa2c2−a4+b2c2−b2a2\Rightarrow b^2c^2\ge qa^2c^2-a^4+b^2c^2-b^2a^2⇒b2c2≥qa2c2−a4+b2c2−b2a2
c2≤a2+b2∴∣c∣≤a2+b2c^2\le a^2+b^2\therefore |c|\le\sqrt{a^2+b^2}c2≤a2+b2∴∣c∣≤a2+b2
cosA1−tanA+sinA1−cotA\dfrac{\cos A}{1-\tan A} + \dfrac{\sin A}{1 - \cot A}1−tanAcosA+1−cotAsinAis equal to
If tanθ=34\tan\theta=\dfrac{3}{4}tanθ=43 and 0<θ,<900,0< \theta, < 90^0,0<θ,<900, then the value of sinθcosθ\sin\theta \cos \thetasinθcosθ is
Evaluate
tanA+secA−1tanA−secA+1;\displaystyle \frac { \tan { A } +\sec { A } -1 }{ \tan { A } -\sec { A } +1 } ;tanA−secA+1tanA+secA−1;
Which of the following is / are the value (S) of the expression?
sin A(1+ tan A) + cos A (1+ cot A) ?
1. sec A + cosec A
2. 2 cosec A ( sin A + cos A )
3. tan A + cot A ;
Select the correct answer using the code given below.& ;