v) Places A and Bare 100 km apart on a highway. One car starts from A and another
from B at the same time. If the cars travel

2 thoughts on “v) Places A and Bare 100 km apart on a highway. One car starts from A and another<br />from B at the same time. If the cars travel”

1. v) Places A and Bare 100 km apart on a highway. One car starts from A and another

   v) Places A and Bare 100 km apart on a highway. One car starts from A and anotherfrom B at the same time. If the cars travel in the same direction at different speeds,

   v) Places A and Bare 100 km apart on a highway. One car starts from A and anotherfrom B at the same time. If the cars travel in the same direction at different speeds,they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What

   v) Places A and Bare 100 km apart on a highway. One car starts from A and anotherfrom B at the same time. If the cars travel in the same direction at different speeds,they meet in 5 hours. If they travel towards each other, they meet in 1 hour. Whatare the speeds of the two cars?

   Reply

2. Answer:

   sinθ+cosθ=a

   secθ+cscθ=b

   \sf\underline \red{ To\:Find}

   ToFind

   We have to find the value of b(a²-1)

   \sf\underline \pink{ Solution }

   Solution

   By putting the given values

   \begin{gathered}:\implies\sf\ \ b(a^2-1)\\ \\ \\ :\implies\sf\ \ sec\theta+csc\theta\big\{(sin\theta+cos\theta)^2-1)\big\}\\ \\ \\ \bullet\sf\ sec\theta=\dfrac{1}{cos\theta}\ \ ;\ csc\theta=\dfrac{1}{sin\theta}\\ \\ \\ :\implies\sf\ \dfrac{1}{cos\theta}+\dfrac{1}{sin\theta}\big\{sin^2\theta+cos^2\theta+2sin\theta cos\theta-1\big\}\\ \\ \\ \bullet\sf\ \ sin^2\theta+cos^2\theta=1\\ \\ \\ :\implies\sf\dfrac{sin\theta+cos\theta}{sin\theta\ cos\theta}\big\{\cancel{1}+2sin\theta\ cos\theta \cancel{-1}\big\}\\ \\ \\ :\implies\sf\ \dfrac{sin\theta+cos\theta}{\cancel{sin\theta cos\theta}}\times 2\cancel{sin\theta cos\theta}\\ \\ \\ :\implies\sf\ \ 2(sin\theta+cos\theta)\end{gathered}

   :⟹ b(a

   2

   −1)

   :⟹ secθ+cscθ{(sinθ+cosθ)

   2

   −1)}

   ∙ secθ=

   cosθ

   1

   ; cscθ=

   sinθ

   1

   :⟹

   cosθ

   1

   +

   sinθ

   1

   {sin

   2

   θ+cos

   2

   θ+2sinθcosθ−1}

   ∙ sin

   2

   θ+cos

   2

   θ=1

   :⟹

   sinθ cosθ

   sinθ+cosθ

   {

   1

   +2sinθ cosθ

   −1

   }

   :⟹

   sinθcosθ

   sinθ+cosθ

   ×2

   sinθcosθ

   :⟹ 2(sinθ+cosθ)

   \underline{\bigstar{\blue{\sf\ \ b(a^2-1)= 2(sin\theta+cos\theta)}}}

   ★ b(a

   2

   −1)=2(sinθ+cosθ

   Reply