Free Introduction to Trigonometry 03 Practice Test - 10th Grade 

We know that, $cos{90}^{\circ }=0$
The given expression
$cos{1}^{\circ }$ $\times$ $cos{2}^{\circ }$ $\times$ $cos{3}^{\circ }$  $\times$....$\times$ $cos{90}^{\circ }$  $\times$……..$\times$ $cos{180}^{\circ }$ reduces to zero as it contains  $cos{90}^{\circ }$ which is equal to 0.

$\frac{sinA-sinB}{cosA+cosB}+\frac{cosA-cosB}{sinA+sinB}$

= $\frac{(sinA-sinB)(sinA+sinB)+(cosA-cosB)(cosA+cosB)}{(cosA+cosB)(sinA+sinB)}$
=$\frac{si{n}^{2}A-si{n}^{2}B+co{s}^{2}A-co{s}^{2}B}{(cosA+cosB)(sinA+sinB)}$
= $\frac{si{n}^{2}A+co{s}^{2}A-(si{n}^{2}B‘+co{s}^{2}B)}{(cosA+cosB)(sinA+sinB)}$
= $\frac{1-1}{(cosA+cosB)(sinA+sinB)}$
= 0

$\frac{\text{tan A}}{1-\text{cot A}}+\frac{\text{cot A}}{1-\text{tan A}}$

$=\frac{\text{tan A}}{1-\left(\frac{1}{\text{tan A}}\right)}+\frac{\frac{1}{\text{tan A}}}{1-\text{tan A}}$

$=\frac{\text{tan A}}{\frac{\text{tan A}-1}{\text{tan A}}}+\frac{1}{\text{tan A}(1-\text{tan A})}$

$=\frac{{\text{tan}}^{2}A}{\text{tan A}-1}-\frac{1}{\text{tan A}(\text{tan A}-1)}\dots \dots (a-b=-(b-a\left)\right)$

$=\frac{{\text{tan}}^{3}A-1}{\text{tan A}(\text{tan A}-1)}$

$=\frac{\left(\text{/}\text{tan A}-1\right)({\text{tan}}^{2}A+\text{tan A}+1)}{\text{tan A}\left(\text{/}\text{tan A}-1\right)}\dots \dots [a^3-b^3=(a-b\left)\right(a^2+ab+b^2\left)\right]$

$=\frac{{\text{tan}}^{2}A}{\text{tan A}}+\frac{\text{tan A}}{\text{tan A}}+\frac{1}{\text{tan A}}$

$=\text{tan A + 1 + cot A}$

$=1+\frac{\text{sin A}}{cosA}+\frac{\text{cos A}}{sinA}$

$=1+\left(\frac{{\text{sin}}^{2}A+{\text{cos}}^{2}A}{\text{sin A}\text{cos A}}\right)$

$=1+\frac{1}{\text{sin A}\text{cos A}}$

$=1+\text{cosec A sec A}$

$\frac{(\text{sec A}+\text{tan A}-1)(\text{sec A}-\text{tan A}+1)}{\text{tan A}}$

$=\frac{[\text{sec A}+(\text{tan A}-1\left)\right][\text{sec A}-(\text{tan A}-1\left)\right]}{\text{tan A}}$

$=\frac{{\text{sec}}^{2}A-(\text{tan A}-1{)}^2}{\text{tan A}}$

$=\frac{{\text{sec}}^{2}A-({\text{tan}}^{2}A-2\text{tan A}+1)}{\text{tan A}}$

$=\frac{{\text{sec}}^{2}A-{\text{tan}}^{2}A+2\text{tan A}-1}{\text{tan A}}$

$=\frac{1+2\text{tan A}-1}{\text{tan A}}\dots \dots (\because 1+{\text{tan}}^{2}A={\text{sec}}^{2}A)$

$=\frac{2\text{tan A}}{\text{tan A}}$

$=2$

$x=a\text{sec}\theta +b\text{tan}\theta ,y=a\text{tan}\theta +b\text{sec}\theta$

Squaring both sides we get,

$x^2=(a\text{sec}\theta +b\text{tan}\theta {)}^2,y^2=(a\text{tan}\theta +b\text{sec}\theta {)}^2$

$\begin{array}{cc}\therefore & x^2=a^2{\text{sec}}^2\theta +2ab\text{sec}\theta \text{tan}\theta +b^2{\text{tan}}^2\theta \\ & y^2=a^2{\text{tan}}^2\theta +2ab\text{sec}\theta \text{tan}\theta +b^2{\text{sec}}^2\theta \\ & \underset{–}{---}\\ \text{Subtracting},& x^2-y^2=a^2({\text{sec}}^2\theta -{\text{tan}}^2\theta )+b^2({\text{tan}}^2\theta -{\text{sec}}^2\theta )\end{array}$

$x^2-y^2=a^2({\text{sec}}^2\theta -{\text{tan}}^2\theta )-b^2{\text{(sec}}^2\theta -{\text{tan}}^2\theta )$

$x^2-y^2=a^2-b^2\dots \dots ({\text{sec}}^2\theta -{\text{tan}}^2\theta =1)$

$\therefore \frac{x^2-y^2}{a^2-b^2}=1$

$\frac{{\text{tan}}^2\theta }{(\text{sec}\theta -1{)}^2}=\frac{{\text{sec}}^2\theta -1}{(\text{sec}\theta -1{)}^2}$

$=\frac{(\text{sec}\theta -1)(\text{sec}\theta +1)}{(sec\theta -1)(\text{sec}\theta -1)}$

$=\frac{\frac{1}{\text{cos}\theta }+1}{\frac{1}{\text{cos}\theta }-1}$

$=\frac{\frac{1+\text{cos}\theta }{\text{cos}\theta }}{\frac{1-\text{cos}\theta }{\text{cos}\theta }}$

$=\frac{1+\text{cos}\theta }{1-\text{cos}\theta }$

$\frac{\text{sec}{17}^{\circ }}{\text{cosec}{73}^{\circ }}+\frac{\text{tan}{68}^{\circ }}{\text{cot}{22}^{\circ }}-[{\text{cos}}^2{44}^{\circ }+{\text{cos}}^2{46}^{\circ }]$

$=\frac{\text{sec}{17}^{\circ }}{\text{cosec}({90}^{\circ }-{17}^{\circ })}+\frac{\text{tan}{68}^{\circ }}{\text{cot}({90}^{\circ }-{68}^{\circ })}-[{\text{cos}}^2{44}^{\circ }+{\text{cos}}^2({90}^{\circ }-{44}^{\circ }\left)\right]$

$=\frac{\text{sec}{17}^{\circ }}{\text{sec}{17}^{\circ }}+\frac{\text{tan}{68}^{\circ }}{\text{tan}{68}^{\circ }}-[{\text{cos}}^2{44}^{\circ }+{\text{sin}}^2{44}^{\circ }]$....(complementary angles)

$=1+1-1$

$=1$

$\Rightarrow (1+\tan A{)}^2+(1-\tan A{)}^2$
$=(1+{\tan }^{2}A+2\tan A)+(1+{\tan }^{2}A-2\tan A)$
$=(2+2{\tan }^{2}A+2\tan A-2\tan A)$
$=(2+2{\tan }^{2}A)$
$=2(1+{\tan }^{2}A)$
$=2{\sec }^{2}A$

We know that,$1+ta{n}^2\theta =se{c}^2\theta \text{and}1+co{t}^2\theta =cose{c}^2\theta$

$\frac{1}{1+ta{n}^2\theta }+\frac{1}{1+co{t}^2\theta }=\frac{1}{se{c}^2\theta }+\frac{1}{cose{c}^2\theta }$

$=co{s}^2\theta +si{n}^2\theta$

$=1$

$\Rightarrow (\sin A+\text{cosec A}{)}^2-(\sin A-\text{cosec A}{)}^2$

$=4\sin A\text{cosec A}$ 

as $\left[\right(a+b{)}^2-(a-b{)}^2=4ab]$

now, $(\sin A\times \text{cosec A}=\sin A\times \frac{1}{\sin A}=1)$

$\therefore 4.\sin A\text{cosec A}=4\times 1=4$