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Chapter 13: Inverse Trigonometric Functions @fsc:fsc_part_1_solutions
8 Hits, Last modified: 5 months ago
\sin ^{ - 1}}A + {\sin ^{ - 1}}B = {\sin ^{ - 1}}\left( {A\sqrt {1 - {B^2}} + B\sqrt {1 - {A^2}} } \rig... \sin ^{ - 1}}A - {\sin ^{ - 1}}B = {\sin ^{ - 1}}\left( {A\sqrt {1 - {B^2}} - B\sqrt {1 - {A^2}} } \rig... \cos ^{ - 1}}A + {\cos ^{ - 1}}B = {\cos ^{ - 1}}\left( {AB - \sqrt {\left( {1 - {A^2}} \right)\left( {1 - {B^2}} \right)} } \right)$ * ${\cos ^{ - 1}}A - {
Chapter 14: Solutions of Trigonometric Equation @fsc:fsc_part_1_solutions
8 Hits, Last modified: 5 months ago
\sin ^{ - 1}}A + {\sin ^{ - 1}}B = {\sin ^{ - 1}}\left( {A\sqrt {1 - {B^2}} + B\sqrt {1 - {A^2}} } \rig... \sin ^{ - 1}}A - {\sin ^{ - 1}}B = {\sin ^{ - 1}}\left( {A\sqrt {1 - {B^2}} - B\sqrt {1 - {A^2}} } \rig... \cos ^{ - 1}}A + {\cos ^{ - 1}}B = {\cos ^{ - 1}}\left( {AB - \sqrt {\left( {1 - {A^2}} \right)\left( {1 - {B^2}} \right)} } \right)$ * ${\cos ^{ - 1}}A - {
Unit 02: Differentiation @fsc:fsc_part_2_solutions
3 Hits, Last modified: 5 months ago
. ===Method 1=== $$ \begin{aligned} \frac{d}{dx}\left(\frac{x+1}{x-1}\right) &= \frac{(x-1)\frac{d}{dx}... (x-1)^{-1} $$ Now $$ \begin{aligned} \frac{d}{dx}\left(\frac{x+1}{x-1}\right) &=\frac{d}{dx}\left(1+2(x-1)^{-1}\right)\\ &= 0-2(x-1)^{-2}(1)\\ &= \frac{-2}{(x
FSc Part 1 (KPK Boards)
1 Hits, Last modified: 5 months ago
define combination and know the notation $^nC_r=\left(\begin{smallmatrix}n\\ r\end{smallmatrix} \right)
FSc Part 2 (KPK Boards)
1 Hits, Last modified: 5 months ago
us and discontinuous functions. * recognize the left hand and right hand limits through examples. *
Chapter 01: Number System @fsc:fsc_part_1_solutions
1 Hits, Last modified: 5 months ago
real and imaginary parts of (i) $(x+iy)^n$ (ii) $\left(\frac{x_1+iy_1}{x_2+iy_2}\right)^n, x_2+iy_2\neq