Question
Question: The value of k, for which \[{\left( {\cos x + \sin x} \right)^2} + k\sin x\cos x - 1 = 0\] is an ide...
The value of k, for which (cosx+sinx)2+ksinxcosx−1=0 is an identity is
(1) −1
(2) −2
(3) 0
(4) 1
Solution
To find the value of k first we have to simplify the given equation. To simplify the equation use a plus b whole square formula for the bracket term. Then a trigonometric identity will be used that is cos2x+sin2x=1 . Then you will find that there are both one and minus one present in the equation, so they both together gives the value as zero. Then take the terms common when needed and solve it further to get the value of k.
Complete step by step solution:
It is given to us that (cosx+sinx)2+ksinxcosx−1=0 is an identity. So, first we will simplify this equation to find the value of k.
⇒(cosx+sinx)2+ksinxcosx−1=0 -------- (i)
By using the formula, (a+b)2=a2+b2+2ab in equation (i) we get,
⇒cos2x+sin2x+2cosxsinx+ksinxcosx−1=0 ------- (ii)
We all know that cos2x+sin2x=1 is a trigonometric identity. So by using this identity the equation (ii) becomes
⇒1+2cosxsinx+ksinxcosx−1=0 ---------- (iii)
Now both ones will cancel out because 1−1=0 .Therefore the equation (iii) becomes
⇒2cosxsinx+ksinxcosx=0 --------- (iv)
Now take out sinxcosx common in the equation (iv) , by this our equation (iv) becomes
⇒sinxcosx(2+k)=0
Now either sinxcosx=0 or (2+k)=0
As we have to find the value of k. Therefore,
⇒2+k=0
By taking 2 to the right hand side we get,
⇒k=−2
**Hence, the correct option is (2) −2
Note:**
Remember the formulas that we used above in the solution part because they are the most easier and commonly used formulas and it is important to remember them. Remember that in the third last step we have two options but we have chosen the second one in the term of k because in the question we are asking for the value of k.