R K Bansal Solutions for Exercise 1: EXERCISE

Prove that: $(1+ \tan \mathrm{A} . \tan \mathrm{B}{)}^2+ (\tan \mathrm{A}- \tan \mathrm{B}{)}^2={\sec }^2 A {\sec }^2 \mathrm{B}$.

Prove that $\frac{1}{\cos \mathrm{A}+\sin \mathrm{A} -1}+\frac{1}{\cos {\displaystyle }{\displaystyle \mathrm{A}}{\displaystyle +}{\displaystyle \sin }{\displaystyle }{\displaystyle \mathrm{A}}{\displaystyle }{\displaystyle +}{\displaystyle 1}}= \mathrm{cosec} \mathrm{A} + \sec \mathrm{A}$.

If$\mathrm{X} \cos \mathrm{A} + \mathrm{y} \sin \mathrm{A} = \mathrm{m}$ and $\mathrm{X} \sin \mathrm{A}- \mathrm{y} \cos \mathrm{A} = \mathrm{n},$ then prove that: ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{m}}^2+{\mathrm{n}}^2$

$\mathrm{If} \mathrm{m} = \mathrm{a} \sec \mathrm{A} + \mathrm{b} \tan \mathrm{A} \mathrm{and}$

$\mathrm{n} = \mathrm{a} \tan \mathrm{A} + \mathrm{b} \sec \mathrm{A},$ then prove that ${\mathrm{m}}^2- {\mathrm{n}}^2= {\mathrm{a}}^2 -{\mathrm{b}}^2$

$\mathrm{If} \mathrm{x} = \mathrm{r} \sin \mathrm{A} \cos \mathrm{B}, \mathrm{y} = \mathrm{r} \sin \mathrm{A} \sin \mathrm{B}$

$\mathrm{z} = \mathrm{r} \cos \mathrm{A},$ then prove that ${\mathrm{x}}^2+ {\mathrm{y}}^2 + {\mathrm{z}}^2={\mathrm{r}}^2$
 

If$\sin \mathrm{A} + \cos \mathrm{A} = \mathrm{m}$ and $\sec \mathrm{A} + \mathrm{cosec} \mathrm{A}= \mathrm{n},$ show that $\mathrm{n}({\mathrm{m}}^2-1)=2 \mathrm{m}$

If$\mathrm{X} = \mathrm{r} \cos \mathrm{A} \cos \mathrm{B}, \mathrm{y} = \mathrm{r} \cos \mathrm{A} \sin \mathrm{B}$and

$\mathrm{z}=\mathrm{r} \sin \mathrm{A},$ show that:

${\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2={\mathrm{r}}^2$

If $\frac{\cos \mathrm{A}}{\cos \mathrm{B}}=\mathrm{m}$  and  $\frac{\cos \mathrm{A}}{\sin \mathrm{B}}=\mathrm{n},$

show that:

$({\mathrm{m}}^2+{\mathrm{n}}^2) {\cos }^2 \mathrm{B}={\mathrm{n}}^2.$