Question

please solve this...,fast...​

Answer 1

Explanation:

hope this may hlp u mate ☺️... answer for 26th qn is this

Answer 2

Question:

1. If $\displaystyle{\sf\:\sin\:\theta\:=\:\dfrac{3}{5}}$, evaluate: $\displaystyle{\sf\:\dfrac{cosec\:\theta\:-\:\cot\:\theta}{2\:\cot\:\theta}}$

2. Find the 20ᵗʰ term from last of the AP 3, 8, 13,......253.

Answer:

1. $\displaystyle{\underline{\boxed{\red{\sf\:\dfrac{cosec\:\theta\:-\:\cot\:\theta}{2\:\cot\:\theta}\:=\:\dfrac{1}{8}}}}}$

2. The 20ᵗʰ term from last of the given AP is 158.

Step-by-step-explanation:

1.

We have given that,

$\displaystyle{\sf\:\sin\:\theta\:=\:\dfrac{3}{5}}$

We have to find the value of,

$\displaystyle{\sf\:\dfrac{cosec\:\theta\:-\:\cot\:\theta}{2\:\cot\:\theta}}$

Now,

$\displaystyle{\sf\:\sin\:\theta\:=\:\dfrac{3}{5}}$

$\displaystyle{\sf\:cosec\:\theta\:=\:\dfrac{1}{\sin\:\theta}\:\:\:-\:-\:[\:Identity\:]}$

$\displaystyle{\implies\sf\:cosec\:\theta\:=\:\dfrac{1}{\dfrac{3}{5}}}$

$\displaystyle{\implies\sf\:cosec\:\theta\:=\:\dfrac{1}{3}\:\times\:\dfrac{5}{1}}$

$\displaystyle{\implies\boxed{\pink{\sf\:cosec\:\theta\:=\:\dfrac{5}{3}}}}$

Now, we know that,

$\displaystyle{\pink{\sf\:1\:+\:\cot^2\:\theta\:=\:cosec^2\:\theta}\sf\:\:\:-\:-\:[\:Trigonometric\:identity\:]}$

$\displaystyle{\implies\sf\:1\:+\:\cot^2\:\theta\:=\:\left(\:\dfrac{5}{3}\:\right)^2}$

$\displaystyle{\implies\sf\:1\:+\:\cot^2\:\theta\:=\:\dfrac{5}{3}\:\times\:\dfrac{5}{3}}$

$\displaystyle{\implies\sf\:1\:+\:\cot^2\:\theta\:=\:\dfrac{25}{9}}$

$\displaystyle{\implies\sf\:\cot^2\:\theta\:=\:\dfrac{25}{9}\:-\:1}$

$\displaystyle{\implies\sf\:\cot^2\:\theta\:=\:\dfrac{25\:-\:9}{9}}$

$\displaystyle{\implies\sf\:\cot^2\:\theta\:=\:\dfrac{16}{9}}$

$\displaystyle{\implies\sf\:\cot\:\theta\:=\:\sqrt{\left(\:\dfrac{16}{9}\:\right)}\:\:\:\:-\:-\:[\:Taking\:square\:roots\:]}$

$\displaystyle{\implies\sf\:\cot\:\theta\:=\:\sqrt{\left(\:\dfrac{4\:\times\:4}{3\:\times\:3}\:\right)}}$

$\displaystyle{\implies\boxed{\red{\sf\:\cot\:\theta\:=\:\dfrac{4}{3}}}}$

Now,

$\displaystyle{\sf\:\dfrac{cosec\:\theta\:-\:\cot\:\theta}{2\:\cot\:\theta}}$

$\displaystyle{\implies\sf\:\dfrac{\dfrac{5}{3}\:-\:\dfrac{4}{3}}{2\:\times\:\dfrac{4}{3}}}$

$\displaystyle{\implies\sf\:\dfrac{\dfrac{5\:-\:4}{3}}{\dfrac{8}{3}}}$

$\displaystyle{\implies\sf\:\dfrac{\dfrac{1}{3}}{\dfrac{8}{3}}}$

$\displaystyle{\implies\sf\:\dfrac{1}{\cancel{3}}\:\times\:\dfrac{\cancel{3}}{8}}$

$\displaystyle{\implies\underline{\boxed{\red{\sf\:\dfrac{cosec\:\theta\:-\:\cot\:\theta}{2\:\cot\:\theta}\:=\:\dfrac{1}{8}}}}}$

──────────────────────────

2.

We have given an AP, 3, 8, 13,......253.

We have to find the 20ᵗʰ term from last of this AP.

Here,

$\displaystyle{\bullet\:\sf\:a\:=\:t_1\:=\:3}$

$\displaystyle{\bullet\:\sf\:d\:=\:t_2\:-\:t_1\:=\:8\:-\:3\:=\:5}$

$\displaystyle{\bullet\:\sf\:t_n\:=\:253}$

Now, we know that,

$\displaystyle{\pink{\sf\:t_n\:=\:a\:+\:(\:n\:-\:1\:)\:\times\:d}\sf\:\:\:-\:-\:[\:Formula\:]}$

$\displaystyle{\implies\sf\:253\:=\:3\:+\:(\:n\:-\:1\:)\:\times\:5}$

$\displaystyle{\implies\sf\:253\:-\:3\:=\:(\:n\:-\:1\:)\:\times\:5}$

$\displaystyle{\implies\sf\:(\:n\:-\:1\:)\:\times\:5\:=\:250}$

$\displaystyle{\implies\sf\:n\:-\:1\:=\:\cancel{\dfrac{250}{5}}}$

$\displaystyle{\implies\sf\:n\:-\:1\:=\:50}$

$\displaystyle{\implies\sf\:n\:=\:50\:+\:1}$

$\displaystyle{\implies\boxed{\red{\sf\:n\:=\:51}}}$

Now,

The 20ᵗʰ term from the last 51ˢᵗ term = 51 - 19

∴ The 20ᵗʰ term from the last 51ˢᵗ term = 32

Now, we know that,

$\displaystyle{\pink{\sf\:t_n\:=\:a\:+\:(\:n\:-\:1\:)\:\times\:d}\sf\:\:\:-\:-\:[\:Formula\:]}$

$\displaystyle{\implies\sf\:t_{32}\:=\:3\:+\:(\:32\:-\:1\:)\:\times\:5}$

$\displaystyle{\implies\sf\:t_{32}\:=\:3\:+\:31\:\times\:5}$

$\displaystyle{\implies\sf\:t_{32}\:=\:3\:+\:155}$

$\displaystyle{\implies\underline{\boxed{\red{\sf\:t_{32}\:=\:158}}}}$

∴ The 20ᵗʰ term from last of the given AP is 158.