Value of the expression 2sinx – cos2x is

Q: Value of the expression 2sinx – cos2x is

(A) greater than -3/2

(B) equal to – 3/2

(C) less than -3/2

(D) none of these

Sol. 2 sinx – cos2x = 2 (sin2x + sinx) – 1

$\large = 2 [(sinx + \frac{1}{2})^2 – \frac{1}{4}] – 1$

$\large = 2(sinx + \frac{1}{2})^2 – \frac{3}{2} \ge -\frac{3}{2}$

⇒ 2 sinx – cos2x ≥ – 3/2

Hence (A), (B) are the correct answers.

If 4 sinA + secA = 0 then tanA equals to

Q: If 4 sinA + secA = 0 then tanA equals to

(A) 4 + √2

(B) – 2 + √3

(C) 2 + 4√3

(D) –2 –√3

Sol.  4sinA + secA = 0

$\large 4 sinA + \frac{1}{cosA} = 0$

$\large \frac{4sinA cosA + 1}{cosA} = 0$

4 sinA cosA + 1 = 0

2(2 sinA cosA ) = -1

2( sin2A ) = -1

⇒ sin2A = -1/2

$\large \frac{2 tanA}{1 + tan^2A} = – \frac{1}{2}$

tanA = – 2 ± √3

Hence (B), (D) are the correct answers.

If Tn = sin^nθ + cos^nθ , then $\frac{T_6 – T_4}{T_6} = m$ holds for values of m satisfying

Q: If Tn = sinnθ + cosnθ , then $\frac{T_6 – T_4}{T_6} = m$ holds for values of m satisfying

(A) $\large m \in [-1 , \frac{1}{3}]$

(B) $\large m \in [0 , \frac{1}{3}]$

(C) $\large m \in [-1 , 0]$

(D) $\large m \in [-1 , -\frac{1}{2}]$

Sol: $\large \frac{T_6 – T_4}{T_6} = \frac{-sin^2 2\theta}{4-3sin^2 2\theta} = m$

$\large sin^2 2\theta = \frac{4 m}{-1 + 3 m}$

$\large 0 \le \frac{4 m}{-1 + 3 m} \le 1$

$\large m \in [-1 , 0]$

Hence (C), (D) are the correct answers

If tanθ = -4/3, then sinθ is

Q: If tanθ = -4/3, then sinθ is

(A) 4/5

(B) –4/5

(C) 3/5

(D) none of these

Sol. The negative value of tanθ implies that θ is lying either in the IInd or IVth quadrant where sinθ takes positive and negative sign respectively.

Hence (A) and (B) are the correct answers.