Question

Unit 1: Distance & Displacement

Q1For one dimensional motion, we often choose the x–axis as the line along which the motion takes place. Then the position of an object at any moment is given by its x- coordinates. If the motion is vertical, as for a dropped object, we usually use y–axis. We need to make distinction between the distance an object has travelled and its displacement.

In context of above lines, answer the following questions:

The below figure represents walking of person 70 m east, then 30 m west.





Q1.1Choose the most appropriate answer for person displacement and total distance for the journey, he covered

100m, 100m

40m, 40m

70m, 40m

40m, 100m

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Answer 1

$\huge\bold{Question :}$

If tan θ = ¹/√7 then , show that $\sf \dfrac{csc^2\theta-sec^2\theta}{csc^2\theta+sec^2\theta}=\dfrac{3}{4}$

$\huge\bold{Solution :}$

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$\sf tan\ \theta=\dfrac{1}{\sqrt{7}}$

$:\to \sf tan^2\theta=\dfrac{1}{(\sqrt{7})^2}$

$:\to \sf \textsf{\textbf{\pink{tan ^\text{2} \boldsymbol \theta\ =\ \dfrac{\text{1}}{\text{7}} }}}\ \; \bigstar$

$\sf \dfrac{1}{cot\ \theta}=\dfrac{1}{\sqrt{7}}$

$:\to \sf cot\ \theta=\sqrt{7}$

$:\to \sf cot^2\theta=(\sqrt{7})^2$

$:\to \sf \textsf{\textbf{\green{cot ^\text{2}\ \boldsymbol \theta \ =\ 7}}}\ \; \bigstar$

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LHS

$:\to \bf \blue{\dfrac{csc^2\theta-sec^2\theta}{csc^2\theta+sec^2\theta}}$

From Trigonometric identities ,

csc²θ = 1 + cot²θ

sec²θ = 1 + tan²θ

$:\to \sf \dfrac{(1+cot^2\theta)-(1+tan^2\theta)}{(1+cot^2\theta)+(1+tan^2\theta)}$

tan²θ = ¹/₇

cot²θ = 7

$:\to \sf \dfrac{(1+7)-(1+\frac{1}{7})}{(1+7)+(1+\frac{1}{7})}$

$:\to\ \sf \dfrac{8-\frac{8}{7}}{8+\frac{8}{7}}$

$:\to\ \sf \dfrac{48}{64}$

$:\to\ \textsf{\textbf{\orange{ \dfrac{\text{3}}{\text{4}} }}}\ \; \bigstar$