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(sin 90^o/cos 45^o) + (1/cosec 30^o) 

= (1/(1/√2)) + 1/2

= √2 + 1/2 = (2√2+1)/2

L.H.S. = sin⁴ θ + cos⁴ θ 

= (sin² θ)² + (cos² θ)² = (sin² θ + cos² θ)² – 2sin² θ cos² θ 

… [ v a² + b² = (a + b)² – 2ab] 

= 1 – 2sin² θ cos² θ 

= R.H.S.

(d) 2 sec θ

\(\frac{cos\,\theta}{1-sin\,\theta}+\frac{cos\,\theta}{1+sin\,\theta}\) = \(\frac{cos\,\theta(1+sin\,\theta)+cos\,\theta(1-sin\,\theta)}{(1-sin\,\theta)(1+sin\,\theta)}\)

= \(\frac{cos\,\theta+cos\,\theta\,sin\,\theta+cos\,\theta-cos\,\theta\,sin\,\theta}{1-sin^2\,\theta}\)

= \(\frac{2cos\,\theta}{cos^2\,\theta}= \frac{2}{cos\,\theta}\) = 2 sec θ

L.H.S = 2(sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ) + 1=0 

= sin⁶ θ + cos⁶ θ

= (sin² θ)³ + (cos² θ)³ = (sin² θ + cos² θ)³ – 3 sin² θ cos² θ (sin2 0 + cos2 0) 

…[••• a³ + b³ = (a + b)³ – 3ab(a + b)] 

= (1)³ – 3 sin² θ cos² θ(1) 

= 1-3 sin² θ cos² θ sin⁴ θ + cos⁴ θ 

= (sin² θ)² + (cos² θ)² = (sin² θ + cos² θ)² – 2 sin² θ cos θ

 …[Y a² + b² = (a + b)² – 2ab] 

= 1-2 sin² θ cos² θ 

L.H.S.= 2(sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ) + 1 

= 2(1-3 sin² θ cos² θ) -3(1 – 2 sin² θ cos² θ) + 1 

= 2-6 sin² θ cos² θ – 3 + 6 sin² θ cos² θ + 1 = c 

= R.H.S.

1. tan² θ - sin² θ = \(\frac{sin^2\,\theta}{cos^2\,\theta}\) - sin² θ

= \(\frac{sin^2\,\theta - sin^2\,\theta\,cos^2\,\theta}{cos^2\,\theta}\)

= \(\frac{sin^2\,\theta(1-cos^2\,\theta)}{cos^2\,\theta}\) = \(\frac{sin^2\,\theta}{cos^2\,\theta}\) x sin² θ = tan² θ sin² θ

2. (1 + cot² θ) ( 1– cos θ) (1 + cos θ) 

= ( 1+ cot² θ) (1– cos² θ)

= \(\bigg(1+\frac{cos^2\,\theta}{sin^2\,\theta}\bigg)(1-cos^2\,\theta)\)

= \(\bigg(\frac{sin^2\,\theta + cos^2\,\theta}{sin^2\,\theta}\bigg) \times sin^2\,\theta\)

(∴ 1 – cos² θ = sin² θ) = 1. (cos² θ + sin² θ =1) 

∴ (c) is the correct option.

(a) 1

Given, sin θ = 1 – sin² θ = cos² θ 

∴ cos² θ + cos⁴ θ = sin θ + sin² θ = 1

L.H.S. = cos⁴ θ – sin⁴ θ + 1 

= (cos² θ)² – (sin² θ)² + 1 

= (cos² θ + sin² θ) c(os² θ – sin² θ) +1 

= (1) (cos² θ – sin² θ) + 1 

= cos² θ + (1 – sin² θ) 

= cos² θ + cos² θ = 2cos² θ 

= R.H.S.

(a) 1

sin⁴ θ + 2 cos² θ \(\bigg(1-\frac{1}{sec^2\,\theta}\bigg)\) + cos⁴ θ

= sin⁴ θ + 2cos² θ (1 cos² θ) + cos⁴ θ 

= sin⁴ θ + 2cos² θ. sin² θ + cos⁴ θ 

= (sin² θ + cos² θ)² = 1² = 1

2cos² θ – 11 cos θ + 5 = 0

∴ 2cos² θ – 10 cos θ – cos θ + 5 = 0 

∴ 2cos θ(cos θ – 5) – 1 (cos θ – 5) = 0 

∴ (cos θ – 5) (2cos θ – 1) = 0 cos θ – 5 = 0 or 2cos θ – 1 = 0 

∴ cos θ = 5 or cos θ = 1/2 

Since, -1 ≤ cos θ ≤ 1 

∴ cos θ = 1/2

5tan θ + 3 = 9sec θ 

∴ 5(sec² θ – 1) + 3 = 9sec θ 

∴ 5sec² θ – 5 + 3 = 9sec θ 

∴ 5sec² θ – 9sec θ – 2 = 0 

∴ 5sec² θ – 10 sec θ + sec θ – 2 = 0 

∴ 5sec θ(sec θ – 2) + 1(sec θ – 2) = 0 

∴ (sec θ – 2) (5sec θ + 1) = 0 

∴ sec θ – 2 = 0 or 5sec θ + 1 = 0 

∴ sec θ = 2 or sec θ = -1/5 

Since sec θ ≥ 1 or sec θ ≤ -1, 

sec θ = 2 

∴ θ = 60° … [ ∵ sec 60° = 2]

2cos² 0 = 3sin θ 

∴ 2(1 – sin² θ) = 3sin θ 

∴ 2 – 2sin² θ = 3sin θ 

∴ 2sin² θ + 3sin 9-2 = θ 

∴ 2sin² θ + 4sin θ – sin θ – 2 = θ 

∴ 2sin θ(sin θ + 2) -1 (sin θ + 2) = θ 

∴ (sin θ + 2) (2sin θ – 1) = 0 

∴ sin θ + 2 = 0 or 2sin θ – 1 = 0 

∴ sin θ = -2 or sin θ = 1/2 

Since, -1 ≤ sin θ ≤ 1 

∴ Sin θ = 1/2 

∴ θ = 30° …[ ∵ sin 30 = 1/2]

\(\frac {sin30° + tan 45° - cosec 45° }{sec30° + cos60° + cot 45°}\)

\(= \cfrac {\frac12 + 1 - \sqrt2}{\frac2{\sqrt 3} + \frac 12 + 1}\)

\(= \cfrac {\frac 32 - \sqrt 2}{\frac 32 + \frac 2{\sqrt3}}\)

\(= \cfrac{\frac{3 - 2\sqrt 2}2}{\frac{3\sqrt 3 + 4} {2\sqrt 3}}\)

\(= \frac {(3 - 2\sqrt 2) (3\sqrt 3 - 4)\sqrt 3}{(3\sqrt 3 + 4)(3\sqrt 3 - 4)}\)

\(= \frac {(9\sqrt 3 - 6\sqrt 6 - 12 + 8\sqrt 2)\sqrt 3}{27 - 16}\)

\(=\frac 1{11} (9\sqrt 3 - 6\sqrt 6 + 8\sqrt2 -12) \sqrt 3\)