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^{2}and has 1200 turns. Calculate self-inductance of the toroid. Assume field to be uniform across the cross section of toroid.b-c/a = tan B/2 - tan C/2 / tan B/2 + tan C/2

^{-1}1/3 = 9/4 sin^{-1}2 root 2/3^{2}-b^{2}/a^{2}+b^{2}let o be a point inside a triangle ABC such that angle OAC=angle OCB= angle OBA= alpha .then prove that

1)cota +cot b + cot c=cot alpha

2) cosec

^{2}a+cosec^{2}b+cosec^{2}c= cosec^{2}alphaQ.2. Prove that ${\mathrm{sin}}^{-1}\left(-\frac{1}{2}\right)+{\mathrm{cos}}^{-1}\left(\frac{-\sqrt{3}}{2}\right)={\mathrm{cos}}^{-1}\left(-\frac{1}{2}\right)$

^{2}tanB=b^{2 }tanA, then prove that the triangle is either right triangle or isosceles.^{ }a(b

^{2+c2}) cosA+b (c^{2}^{+a2})cosB+c(^{a2+b2})cosC=3abcQ.5. Find the number of values of (x, y, z) satisfying ${\mathrm{sin}}^{-1}x+{\mathrm{sin}}^{-1}y+{\mathrm{sin}}^{-1}z=\frac{3\mathrm{\pi}}{2}$

Q.6) Prove that ${\mathrm{tan}}^{-1}\left(\sqrt{x}\right)=\frac{1}{2}{\mathrm{cos}}^{-1}\left(\frac{1-x}{1+x}\right)ifx\in \left[0,1\right]$

How many states in India?Q.4) If ${\mathrm{tan}}^{-1}\left(\frac{3{a}^{2}x-{x}^{3}}{{a}^{3}-3a{x}^{2}}\right)=k\mathrm{tan}\left(\frac{x}{a}\right)$. Find k.

^{2}+ BC^{2 }= AC^{2}$Prove\frac{{\mathrm{tan}}^{2}\theta}{sec\theta +1}=\frac{1+\mathrm{cos}\theta}{1-\mathrm{cos}\theta}$

(1) $2p\left(\mathrm{or}\right)\frac{1}{2p}$ (2) $\frac{1}{2p}\left(\mathrm{or}\right)-2p$ (3) $-\frac{1}{2p}\left(\mathrm{or}\right)2p$ (4) $-\frac{1}{2p}\left(\mathrm{or}\right)-2p$

^{2}+b^{2}c^{2}=3(OA^{2}+OB^{2}+OC^{2})- 9 OG^{2}where O is any point in plane of triangle ABC....kindly give proof using vector algebra rather than plane geometry (i know all concepts related to vector analysis and vector algebra)a

^{2}sin(B-C)/sin A + b^{2}sin(C-A)/sin B + c^{2}sin(A-b)/sin C = 0quick ans neededthank u

Q. Prove:- $\frac{{\mathrm{tan}}^{2}\mathrm{\theta}}{\mathrm{sec}\mathrm{\theta}+1}$ = $\frac{1+\mathrm{cos}\mathrm{\theta}}{1-\mathrm{cos}\mathrm{\theta}}$

$3.\mathrm{If}\frac{\mathrm{sin}\left(\mathrm{\theta}+\mathrm{A}\right)}{\mathrm{sin}\left(\mathrm{\theta}+\mathrm{B}\right)}=\sqrt{\frac{\mathrm{sin}2\mathrm{A}}{\mathrm{sin}2\mathrm{B}}}\mathrm{then}{\mathrm{tan}}^{2}\mathrm{\theta}=\phantom{\rule{0ex}{0ex}}\left(1\right)\mathrm{tan}\mathrm{A}\left(2\right)\mathrm{tan}\mathrm{B}\left(3\right)\mathrm{tan}\mathrm{A}+\mathrm{tan}\mathrm{B}\left(4\right)\mathrm{tan}\mathrm{A}\mathrm{tan}\mathrm{B}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}4.\mathrm{If}\mathrm{tan}\mathrm{\beta}=2\mathrm{sin}\mathrm{\alpha}\mathrm{sin}\mathrm{\gamma}\mathrm{cos}\mathrm{ec}\left(\mathrm{\alpha}+\mathrm{\gamma}\right),\mathrm{then}\mathrm{cot}\mathrm{\alpha},\mathrm{cot}\mathrm{\beta}\mathrm{and}\mathrm{cot}\mathrm{\gamma}\mathrm{are}\mathrm{in}.\phantom{\rule{0ex}{0ex}}\left(1\right)\mathrm{A}.\mathrm{P}.\left(2\right)\mathrm{G}.\mathrm{P}.\left(3\right)\mathrm{H}.\mathrm{P}.\left(4\right)\mathrm{A}.\mathrm{G}.\mathrm{P}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}5\mathrm{If}\mathrm{tan}\mathrm{\beta}=\frac{\mathrm{n}\mathrm{tan}\mathrm{\alpha}}{1+\left(1-\mathrm{n}\right){\mathrm{tan}}^{2}\mathrm{\alpha}}\mathrm{tan}\left(\mathrm{\alpha}-\mathrm{\beta}\right)=\phantom{\rule{0ex}{0ex}}\left(1\right)\left(1+\mathrm{n}\right)\mathrm{tan}\mathrm{\alpha}\left(2\right)\left(1-\mathrm{n}\right)\mathrm{tan}\mathrm{\alpha}\phantom{\rule{0ex}{0ex}}\left(3\right)-\left(1+\mathrm{n}\right)\mathrm{tan}\mathrm{\alpha}\left(4\right)-\left(1-\mathrm{n}\right)\mathrm{tan}\mathrm{\alpha}$

Dear vivek Kumar mehta please don't answer if you don't know answer because of u I am posting this question for third time

U f**** don't waste chance of others to answer this question u m*d understood I u know then only answer or I will serious actions

Q. Prove :-

$\frac{{\mathrm{tan}}^{2}\mathrm{\theta}}{\mathrm{sec}\mathrm{\theta}+1}=\frac{1+\mathrm{cos}\mathrm{\theta}}{1-\mathrm{cos}\mathrm{\theta}}$