বিপরীত ত্রিকোণমিতিক ফাংশনের যোগ বিয়োগ
tan−1x+tan−1y=?\tan ^{-1} x+\tan ^{-1} y = ? tan−1x+tan−1y=?
12cos−12(x+y)(1−xy)(1+x2)(1+y2)\frac{1}{2} \cos^{-1} \frac{2(x+y)(1-xy)}{\left(1+x^{2}\right)\left(1+y^{2}\right)}21cos−1(1+x2)(1+y2)2(x+y)(1−xy)
12tan−12(x+y)(1−xy)(1+x2)(1+y2)\frac{1}{2} \tan^{-1} \frac{2(x+y)(1-xy)}{\left(1+x^{2}\right)\left(1+y^{2}\right)}21tan−1(1+x2)(1+y2)2(x+y)(1−xy)
12sin−12(x+y)(1−xy)(1+x2)(1+y2)\frac{1}{2} \sin^{-1} \frac{2(x+y)(1-xy)}{\left(1+x^{2}\right)\left(1+y^{2}\right)}21sin−1(1+x2)(1+y2)2(x+y)(1−xy)
12sec−12(x+y)(1−xy)(1+x2)(1+y2)\frac{1}{2} \sec^{-1} \frac{2(x+y)(1-xy)}{\left(1+x^{2}\right)\left(1+y^{2}\right)}21sec−1(1+x2)(1+y2)2(x+y)(1−xy)
Solve:
tan−1x+tan−1y=tan−1x+y1−xy=12⋅2tan−1x+y1−xy=12sin−12x+y1−xy1+(x+y1−xy)2=12sin−12x+y1−xy1−2xy+x2y2+x2+2xy+y2(1−xy)2=12sin−12(x+y)(1−xy)1+x2y2+x2+y2=12sin−12(x+y)(1−xy)(1+x2)(1+y2)\begin{array}{l}\tan ^{-1} x+\tan ^{-1} y \\=\tan ^{-1} \frac{x+y}{1-x y}=\frac{1}{2} \cdot 2 \tan ^{-1} \frac{x+y}{1-x y} \\=\frac{1}{2} \sin ^{-1} \frac{2 \frac{x+y}{1-x y}}{1+\left(\frac{x+y}{1-x y}\right)^{2}} \\=\frac{1}{2} \sin ^{-1} \frac{2 \frac{x+y}{1-x y}}{\frac{1-2 x y+x^{2} y^{2}+x^{2}+2 x y+y^{2}}{(1-x y)^{2}}} \\=\frac{1}{2} \sin ^{-1} \frac{2(x+y)(1-x y)}{1+x^{2} y^{2}+x^{2}+y^{2}} \\=\frac{1}{2} \sin ^{-1} \frac{2(x+y)(1-x y)}{\left(1+x^{2}\right)\left(1+y^{2}\right)}\end{array}tan−1x+tan−1y=tan−11−xyx+y=21⋅2tan−11−xyx+y=21sin−11+(1−xyx+y)221−xyx+y=21sin−1(1−xy)21−2xy+x2y2+x2+2xy+y221−xyx+y=21sin−11+x2y2+x2+y22(x+y)(1−xy)=21sin−1(1+x2)(1+y2)2(x+y)(1−xy)
Ai এর মাধ্যমে
১০ লক্ষ+ প্রশ্ন ডাটাবেজ
প্র্যাকটিস এর মাধ্যমে নিজেকে তৈরি করে ফেলো
উত্তর দিবে তোমার বই থেকে ও তোমার মত করে।
সারা দেশের শিক্ষার্থীদের মধ্যে নিজের অবস্থান যাচাই
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.
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
মান নির্ণয় কর:
sin−1(4/5)+sin−1(5/13)+sin−1(16/65)\sin^{-1}(4/5)+\sin^{-1}(5/13)+\sin^{-1}(16/65)sin−1(4/5)+sin−1(5/13)+sin−1(16/65)
