Question

If 3 sec (3x-21) = 2 , then the value of Sin²(x +13) + cot^2(x+13)​

Answer 1

Answer:

Find the value of x in the following:

2sin3x=

3

Medium

Solution

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Given,

2sin3x=

3

sin3x=

2

3

we know that,

sin60

∘

=

2

3

∴sin3x=sin60

∘

3x=60

∘

x=

3

60

∘

x=20

∘

Answer 2

Answer:

Correct question is: $\sqrt{3} sec (3x-21)\textdegree = 2$, then the value of $sin^2(x +13)\textdegree + cot^2(x+13)\textdegree$

Final Answer:$sin^2(x +13)\textdegree + cot^2(x+13)\textdegree=\frac{13}{4}$

Step-by-step explanation:

The given question is, $\sqrt{3} sec (3x-21)\textdegree = 2$ using this one we should find the value of $x$.

             $\sqrt{3} sec (3x-21)\textdegree = 2$

                  $sec (3x-21)\textdegree = \frac{2}{\sqrt{3}}$

In trigonometric table, $sec 30\textdegree=\frac{2}{\sqrt{3}}$. Equate this value to the above equation.

                   $sec (3x-21) \textdegree= \frac{2}{\sqrt{3}}=sec30\textdegree$

                   $sec (3x-21)\textdegree =sec30\textdegree$

It can be written as,

                             $3x-21 =30$

                                     $3x=30+21$

                                      $3x=51$

                                        $x=\frac{51}{3}$

                                        $x=17$

Now we can solve $sin^2(x +13) \textdegree+ cot^2(x+13)\textdegree$

                                            $=sin^2(x +13)\textdegree + cot^2(x+13)\textdegree$

                                            $=sin^2(17 +13)\textdegree + cot^2(17+13)\textdegree$

                                            $=sin^230\textdegree + cot^230\textdegree$

                                            $=(sin30\textdegree)^2 + (cot30\textdegree)^2$

In trigonometric table, $sin 30\textdegree=\frac{1}{2}$ and $cot 30\textdegree=\sqrt{3}$

                                             $=(\frac{1}{2} )^2+(\sqrt{3} )^2$    ⇒    $(\frac{1}{2})^2=\frac{1}{4}$ and $(\sqrt{3})^2=3$

                                             $=\frac{1}{4}+3$

                                              $=\frac{1+12}{4}$

$sin^2(x +13) \textdegree+ cot^2(x+13)\textdegree=\frac{13}{4}$