বিপরীত ত্রিকোণমিতিক ফাংশনের যোগ বিয়োগ
দৃশ্যকল্প-১: f(θ)=sinθ \mathbf{f}(\theta)=\sin \theta f(θ)=sinθ.
দৃশ্যকল্প-২: A=cosec−15−12sin−135+tan−114 A=\operatorname{cosec}^{-1} \sqrt{5}-\frac{1}{2} \sin ^{-1} \frac{3}{5}+\tan ^{-1} \frac{1}{4} A=cosec−15−21sin−153+tan−141.
দেখাও যে, sec2(tan−115)+cosec2(cot−113)=30 \sec ^{2}\left(\tan ^{-1} \sqrt{15}\right)+\operatorname{cosec}^{2}\left(\cot ^{-1} \sqrt{13}\right)=30 sec2(tan−115)+cosec2(cot−113)=30.
দৃশ্যকল্প-১ এর আলোকে 2f(π2−θ)⋅f(π2−3θ)+1=0 2 f\left(\frac{\pi}{2}-\theta\right) \cdot f\left(\frac{\pi}{2}-3 \theta\right)+1=0 2f(2π−θ)⋅f(2π−3θ)+1=0 সমীকরণের সমাধান কর।
দৃশ্যকল্প-২ থেকে দেখাও যে, A=tan−11127 A=\tan ^{-1} \frac{11}{27} A=tan−12711.
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
