বিপরীত ত্রিকোণমিতিক ফাংশনের যোগ বিয়োগ
A=sec−15,B=12sin−1pq,C=sin−1rf(x)=sinαx,g(x)=sinβx. \begin{array}{l}A=\sec ^{-1} \sqrt{5}, B=\frac{1}{2} \sin ^{-1} \frac{p}{q}, C=\sin ^{-1} r \\ f(x)=\sin \alpha x, \quad g(x)=\sin \beta x .\end{array} A=sec−15,B=21sin−1qp,C=sin−1rf(x)=sinαx,g(x)=sinβx.
দেখাও যে, 2tan−1x=tan−12x1−x2 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}} 2tan−1x=tan−11−x22x
p=3,q=5,r=110 p=3, q=5, r=\frac{1}{\sqrt{10}} p=3,q=5,r=101 হলে, প্রমাণ কর যে, A−B+C=cot−1(12) A-B+C=\cot ^{-1}\left(\frac{1}{2}\right) A−B+C=cot−1(21)
α=1,β=3 \alpha=1, \beta=3 α=1,β=3 হলে, −π -\pi −π হতে π \pi π ব্যবধির মধ্যে 2f(x)⋅g(x)=1 2 f(x) \cdot g(x)=1 2f(x)⋅g(x)=1 সমীকরণের সমাধান নির্ণয় কর।
উদ্দীপক-১: f(x)=cosxf(x)=\cos xf(x)=cosx
উদ্দীপক-2: cot−1(1x)+12sec−1(1+y21−y2)+12cosec−1(1+z22z)=π\cot ^{-1}\left(\frac{1}{x}\right)+\frac{1}{2} \sec ^{-1}\left(\frac{1+y^{2}}{1-y^{2}}\right)+\frac{1}{2} \operatorname{cosec}^{-1}\left(\frac{1+z^{2}}{2 z}\right)=\picot−1(x1)+21sec−1(1−y21+y2)+21cosec−1(2z1+z2)=π.
costan−1sincot−1(x)=? \cos \tan ^{-1} \sin \cot ^{-1}(\mathrm{x})=? costan−1sincot−1(x)=?
f(x)=cot(π2−x) এবং g(x)=sin−1x f(x)=\cot \left(\frac{\pi}{2}-x\right) \text { এবং } g(x)=\sin ^{-1} x f(x)=cot(2π−x) এবং g(x)=sin−1x
f(x)=sin−1p+sin−1q+sin−1rA=cosx−cos2xR=1−cosx \begin{array}{l}f(x)=\sin ^{-1} p+\sin ^{-1} q+\sin ^{-1} r \\ A=\cos x-\cos 2 x \\ R=1-\cos x\end{array} f(x)=sin−1p+sin−1q+sin−1rA=cosx−cos2xR=1−cosx
