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if  2\, cos\, \theta +sin\, \theta =1\left ( \theta \neq \frac{\pi }{2} \right ),  then 7\, cos\, \theta +\, 6\, sin\, \theta is equal to :

  • Option 1)

    \frac{1}{2}

  • Option 2)

    2

  • Option 3)

    \frac{11}{2}

  • Option 4)

    \frac{46}{5}

 

Answers (1)

best_answer

As we have learned

Trigonometric Identities -

\sin ^{2}\Theta + \cos ^{2}\Theta = 1

1 + \tan ^{2}\Theta = \sec ^{2}\Theta

1 + \cot ^{2}\Theta = cosec ^{2}\Theta

- wherein

They are true for all real values of \Theta

 

 2 \cos \theta + \sin \theta = 1

\Rightarrow \sin \theta = 1 - 2 \cos \theta \\ \Rightarrow \sin ^2 \theta = 1 + 4 \cos ^2 \theta - 4 \cos \theta \\ \Rightarrow 1-\cos ^2 \theta = 1 + 4 \cos ^2 \theta - 4 \cos \theta \\ 5 \cos ^ 2\theta - 4 \cos \theta = 0 \\ \Rightarrow \cos \theta = 0 , 4/5

But \theta \neq \pi / 2 \Rightarrow \cos \theta = 4/5 \\ Now \: \: \sin \theta = 1- 4 \times 2/5 = -3 / 5

 

\therefore 7 \cos \theta + 6 \sin \theta \\ 7 \times 4/5 + 6 (-3/5 ) = 2

 

 

 

 

 

 

 


Option 1)

\frac{1}{2}

Option 2)

2

Option 3)

\frac{11}{2}

Option 4)

\frac{46}{5}

Posted by

Himanshu

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