(a)

\begin{align} \sec x \cos x & = \left(1 \over \cos x \right) \cos x \\ & = {\cos x \over \cos x} \\ & = 1 \end{align}

(b)

\begin{align} \require{cancel} \sin x \cot x & = \sin x \left(\cos x \over \sin x\right) \\ & = {\cancel{\sin x} \cos x \over \cancel{\sin x}} \\ & = \cos x \end{align}

(c)

\begin{align} (\sec x + 1)(\sec x - 1) & = (\sec x)^2 - (1)^2 \phantom{000000000} [ (a + b)(a - b) = a^2 - b^2] \\ & = \sec^2 x - 1 \\ & = (\tan^2 x + 1) - 1 \phantom{00000000} \text{[Identity: } \tan^2 A + 1 = \sec^2 A] \\ & = \tan^2 x + 1 - 1 \\ & = \tan^2 x \end{align}

(d)

\begin{align} {\sin x \over (1 - \cos x)(1 + \cos x)} & = {\sin x \over (1)^2 - (\cos x)^2} \phantom{00000000} [(a - b)(a + b) = a^2 - b^2] \\ & = {\sin x \over 1 - \cos^2 x} \\ & = {\sin x \over \sin^2 x} \phantom{000000000000000.} [\sin^2 A + \cos^2 A = 1 \implies \sin^2 A = 1 - \cos^2 A ] \\ & = {1 \over \sin x} \\ & = \text{cosec} x \end{align}

(i)

\begin{align} \sin \theta & = -{1 \over 3} \phantom{000000} [\text{3rd or 4th quadrant}] \\ {Opp \over Hyp} & = {-1 \over 3} \end{align}

\begin{align} x & = \pm \sqrt{ (3)^2 - (1)^2} \\ & = \pm \sqrt{8} \\ \\ \sec \theta & = {1 \over \cos \theta} \\ & = {1 \over {Adj \over Hyp}} \\ & = {1 \over {\pm \sqrt{8} \over 3}} \\ & = \pm {3 \over \sqrt{8}} \end{align}

(ii)

\begin{align} \tan \theta & = {Opp \over Adj} \\ & = {1 \over \pm \sqrt{8}} \\ & = \pm {1 \over \sqrt{8}} \end{align}

(a)

\begin{align} \text{L.H.S} & = \sec x \sin x \\ & = \left(1 \over \cos x\right) \sin x \\ & = {\sin x \over \cos x} \\ & = \tan x \\ & = \text{R.H.S} \end{align}

(b)

\begin{align} \require{cancel} \text{L.H.S} & = \cos x \tan x \\ & = \cos x \left(\sin x \over \cos x \right) \\ & = {\sin x \cancel{\cos x} \over \cancel{\cos x}} \\ & = \sin x \\ & = \text{R.H.S} \end{align}

(c)

\begin{align} \text{L.H.S} & = {\cos^2 x + \sin^2 x \over 1 - \cos^2 x} \\ & = {1 \over 1 - \cos^2 x} \phantom{00000} [ \sin^2 A + \cos^2 A = 1 ] \\ & = {1 \over \sin^2 x} \phantom{000000000} [\sin^2 A + \cos^2 A = 1 \implies \sin^2 A = 1 - \cos^2 A] \\ & = \left(1 \over \sin x\right)^2 \\ & = (\text{cosec } x)^2 \\ & = \text{cosec}^2 x \\ & = \text{R.H.S} \end{align}

(d)

\begin{align} \text{L.H.S} & = 2 + 3\sin^2 x \\ & = 2 + 3(1 - \cos^2 x) \phantom{00000000} [\sin^2 A + \cos^2 A = 1 \implies \sin^2 A = 1 - \cos^2 A ] \\ & = 2 + 3 - 3\cos^2 x \\ & = 5 - 3\cos^2 x \\ & = \text{R.H.S} \end{align}

(e)

\begin{align} \text{L.H.S} & = \sin^2 x (1 + \tan^2 x) \\ & = \sin^2 x \left(1 + {\sin^2 x \over \cos^2 x} \right) \\ & = \sin^2 x \left( {\cos^2 x \over \cos^2 x} + {\sin^2 x \over \cos^2 x} \right) \\ & = \sin^2 x \left( {\cos^2 x + \sin^2 x \over \cos^2 x} \right) \\ & = \sin^2 x \left(1 \over \cos^2 x \right) \phantom{0000000000} [\sin^2 A + \cos^2 A = 1 ] \\ & = {\sin^2 x \over \cos^2 x} \\ & = \tan^2 x \\ & = \text{R.H.S} \end{align}

(f)

\begin{align} \text{L.H.S} & = (1 + \cot^2 x) \cos x \\ & = (\text{cosec}^2 x)\cos x \phantom{000000000000} [\cot^2 A + 1 = \text{cosec}^2 A] \\ & = (\text{cosec } x) (\text{cosec } x) \cos x \\ & = (\text{cosec } x)\left(1 \over \sin x\right)\cos x \\ & = \text{cosec } x \left(\cos x \over \sin x\right) \\ & = \text{cosec } x \cot x \\ & = \text{R.H.S} \end{align}

\begin{align} \text{L.H.S} & = (\sin \theta + \cos \theta)^2 \\ & = (\sin \theta)^2 + 2(\sin \theta)(\cos \theta) + (\cos \theta)^2 \\ & = \sin^2 \theta + 2 \sin \theta \cos \theta + \cos^2 \theta \\ & = (\sin^2 \theta + \cos^2 \theta) + 2\sin \theta \cos \theta \\ & = 1 + 2\sin \theta \cos \theta \phantom{0000000000000000} [\text{Identity: } \sin^2 A + \cos^2 A = 1] \\ & = \text{R.H.S} \end{align}

\begin{align} \sin \theta \left( {2 \over 1 - \cos \theta} - {2 \over 1 + \cos \theta} \right) & = \sin \theta \left[ {2(1 + \cos \theta) \over (1 - \cos \theta)(1 + \cos \theta)} - {2(1 - \cos \theta) \over (1 + \cos \theta)(1 - \cos \theta) } \right] \\ & = \sin \theta \left[ {2(1 + \cos \theta) - 2(1 - \cos \theta) \over (1 + \cos \theta)(1 - \cos \theta) } \right] \\ & = \sin \theta \left[ { 2 + 2\cos \theta - 2 + 2 \cos \theta \over (1)^2 - (\cos \theta)^2 } \right] \\ & = \sin \theta \left( {4 \cos \theta \over 1 - \cos^2 \theta} \right) \\ & = \sin \theta \left( {4 \cos \theta \over \sin^2 \theta} \right) \phantom{000000} [\sin^2 A + \cos^2 A = 1 \implies \sin^2 A = 1 - \cos^2 A] \\ & = {4 \sin \theta \cos \theta \over \sin^2 \theta} \\ & = {4 \cos \theta \over \sin \theta} \\ & = 4 \left(\cos \theta \over \sin \theta\right) \\ & = 4 \cot \theta \text{ (Shown)} \end{align}

(i)

\begin{align} x & = 3 \cos \theta \\ {x \over 3} & = \cos \theta \\ \\ y & = 2 \tan \theta \\ {y \over 2} & = \tan \theta \\ {y \over 2} & = {\sin \theta \over \cos \theta} \\ {y \over 2} \cos \theta & = \sin \theta \\ \\ \text{Since } & \cos \theta = {x \over 3}, \\ {y \over 2} \left(x \over 3\right) & = \sin \theta \\ {xy \over 6} & = \sin \theta \\ \\ \therefore \sin \theta & = {xy \over 6} \\ & = {1 \over 6} xy \end{align}

(ii)

\begin{align} \text{Since } & \sin \theta = {1 \over 6}xy, \\ 6 \sin \theta & = xy \\ \\ 4x^2 + x^2y^2 & = 4x^2 + (xy)^2 \\ & = 4(3 \cos \theta)^2 + (6 \sin \theta)^2 \\ & = 4(9 \cos^2 \theta) + 36 \sin^2 \theta \\ & = 36\cos^2 \theta + 36\sin^2 \theta \\ & = 36(\cos^2 \theta + \sin^2 \theta) \\ & = 36(1) \phantom{000000000000000} [\sin^2 A + \cos^2 A = 1] \\ & = 36 \end{align}

(a)

\begin{align} \text{L.H.S} & = \sin^4 x - \cos^4 x \\ & = (\sin^2 x)^2 - (\cos^2 x)^2 \\ & = (\sin^2 x + \cos^2 x)(\sin^2 x - \cos^2 x) \phantom{00000000} [a^2 - b^2 = (a + b)(a - b)] \\ & = (1)(\sin^2 x - \cos^2 x) \phantom{0000000000000000000} [\sin^2 A + \cos^2 A = 1] \\ & = \sin^2 x - \cos^2 x \\ & = \text{R.H.S} \end{align}

(b)

\begin{align} \text{L.H.S} & = \sin^2 x + \tan^2 x \sin^2 x \\ & = \sin^2 x (1 + \tan^2 x) \\ & = \sin^2 x (\sec^2 x) \phantom{0000000000} [\tan^2 A + 1 = \sec^2 A] \\ & = \sin^2 x \left(1 \over \cos^2 x\right) \\ & = {\sin^2 x \over \cos^2 x} \\ & = \tan^2 x \\ & = \text{R.H.S} \end{align}

(c)

\begin{align} \text{L.H.S} & = (1 + \tan x - \sec x)(1 + \cot x + \text{cosec } x) \\ & = 1 + \cot x + \text{cosec } x + \tan x + (\tan x)(\cot x) + (\tan x)(\text{cosec } x) \\ & \phantom{...} - \sec x - (\sec x)(\cot x) - (\sec x)(\text{cosec } x) \\ & = 1 + \cot x + \text{cosec } x + \tan x + (\tan x)\left(1 \over \tan x\right) + \left(\sin x \over \cos x\right)\left(1 \over \sin x\right) \\ & \phantom{...} - \sec x - \left(1 \over \cos x\right)\left(\cos x \over \sin x\right) - \left(1 \over \cos x\right)\left(1 \over \sin x\right) \\ & = 1 + \cot x + \text{cosec } x + \tan x + 1 + {1 \over \cos x} - \sec x - {1 \over \sin x} - {1 \over \sin x \cos x} \\ & = 1 + \cot x + \text{cosec } x + \tan x + 1 + \sec x - \sec x - \text{cosec } x - {1 \over \sin x \cos x} \\ & = 2 + \cot x + \tan x - {1 \over \sin x \cos x} \\ & = 2 + {\cos x \over \sin x} + {\sin x \over \cos x} - {1 \over \sin x \cos x} \\ & = 2 + {\cos^2 x \over \sin x \cos x} + {\sin^2 x \over \sin x \cos x} - {1 \over \sin x \cos x } \\ & = 2 + {(\cos^2 x + \sin^2 x) - 1 \over \sin x \cos x} \\ & = 2 + {1 - 1 \over \sin x \cos x} \phantom{0000000000000} [\sin^2 A + \cos^2 A = 1] \\ & = 2 + 0 \\ & = 2 \\ & = \text{R.H.S} \end{align}

(d)

\begin{align} \text{L.H.S} & = {\sin x \over \sec x - 1} + {\sin x \over \sec x + 1} \\ & = {\sin x(\sec x + 1) \over (\sec x - 1)(\sec x + 1)} + {\sin x(\sec x - 1) \over (\sec x - 1)(\sec x +1)} \\ & = {\sin x(\sec x + 1) + \sin x (\sec x - 1) \over (\sec x - 1)(\sec x + 1)} \\ & = {\sin x \sec x + \sin x + \sin x \sec x - \sin x \over \sec^2 x - 1} \\ & = {2\sin x \sec x \over \sec^2 x - 1} \\ & = {2\sin x \left(1 \over \cos x \right) \over \sec^2 x - 1} \\ & = {2 \left(\sin x \over \cos x\right) \over \sec^2 x - 1} \\ & = {2\tan x \over \sec^2 x - 1} \\ & = {2\tan x \over \tan^2 x} \phantom{000000000} [\tan^2 A + 1 = \sec^2 A \implies \tan^2 A = \sec^2 A - 1] \\ & = {2 \over \tan x} \\ & = 2 \left(1 \over \tan x\right) \\ & = 2\cot x \\ & = \text{R.H.S} \end{align}

(e)

\begin{align} \require{cancel} \text{L.H.S} & = {3 - 6\cos^2 x \over \sin x - \cos x} \\ & = {3(1) - 6\cos^2 x \over \sin x - \cos x} \\ & = {3(\sin^2 x + \cos^2 x) - 6\cos^2 x \over \sin x - \cos x} \phantom{00000000} [\sin^2 A + \cos^2 A = 1 ] \\ & = {3\sin^2 x + 3\cos^2 x - 6\cos^2 x \over \sin x - \cos x} \\ & = {3\sin^2 x - 3\cos^2 x \over \sin x - \cos x} \\ & = {3(\sin^2 x - \cos^2 x) \over \sin x - \cos x} \\ & = {3(\sin x + \cos x)\cancel{(\sin x - \cos x)} \over \cancel{\sin x - \cos x}} \phantom{0000000} [a^2 - b^2 = (a + b)(a - b)] \\ & = 3(\sin x + \cos x) \\ & = \text{R.H.S} \end{align}

(f)

\begin{align} \require{cancel} \text{L.H.S} & = {1 \over 1 - {1 \over 1 - {1 \over 1 - \sec^2 x}}} \\ & = {1 \over 1 - {1 \over 1 - {1 \over 1 - (\tan^2 x + 1)}}} \phantom{00000000} [\tan^2 A + 1 = \sec^2 A] \\ & = {1 \over 1 - {1 \over 1 - {1 \over -\tan^2 x}}} \\ & = {1 \over 1 - {1 \over 1 - (-\cot^2 x)} } \\ & = {1 \over 1 - {1 \over 1 + \cot^2 x}} \\ & = {1 \over 1 - {1 \over \text{cosec}^2 x}} \phantom{000000000000} [\cot^2 A + 1 = \text{cosec}^2 A] \\ & = {1 \over 1 - {1 \over {1 \over \sin^2 x}}} \\ & = {1 \over 1 - \sin^2 x} \\ & = {1 \over \cos^2 x} \phantom{0000000000000} [\sin^2 A + \cos^2 A = 1 \implies \cos^2 A = 1 - \sin^2 A] \\ & = \sec^2 x \\ & = \text{R.H.S} \end{align}

(i)

\begin{align} \text{L.H.S} & = \cot x + \tan x \\ & = {\cos x \over \sin x} + {\sin x \over \cos x} \\ & = {\cos x (\cos x) \over \sin x \cos x} + {\sin x (\sin x) \over \sin x \cos x} \\ & = {\cos^2 x + \sin^2 x \over \sin x \cos x} \\ & = {1 \over \sin x \cos x} \phantom{0000000000} [\sin^2 A + \cos^2 A = 1] \\ & = \left(1 \over \sin x\right)\left(1 \over \cos x\right) \\ & = \text{cosec } x \sec x \\ & = \text{R.H.S} \end{align}

(ii)

\begin{align} \text{L.H.S} & = \text{cosec}^2 x + \sec^2 x \\ & = {1 \over \sin^2 x} + {1 \over \cos^2 x} \\ & = {\cos^2 x \over \sin^2 x \cos^2 x} + {\sin^2 x \over \sin^2 x \cos^2 x} \\ & = {\cos^2 x + \sin^2 x \over \sin^2 x \cos^2 x} \\ & = {1 \over \sin^2 x \cos^2 x} \phantom{0000000000} [\sin^2 A + \cos^2 A = 1] \\ & = \left(1 \over \sin^2 x\right) \left(1 \over \cos^2 x\right) \\ & = \text{cosec}^2 x \sec^2 x \\ & = \text{R.H.S} \end{align}

Hence

\begin{align} \text{L.H.S} & = (\cot x + \tan x)^2 \\ & = (\text{cosec } x \sec x)^2 \phantom{000000000} \text{[Apply result from part i]} \\ & = \text{cosec}^2 x \sec^2 x \\ & = \text{cosec}^2 x + \sec^2 x \phantom{0000000} \text{[Apply result from part ii]} \\ & = \text{R.H.S} \end{align}

\begin{align} \text{L.H.S} & = (\sec x + \tan x)(\sec x - \tan x) + (\text{cosec } x + \cot x)(\text{cosec } x - \cot x) \\ & = (\sec x)^2 - (\tan x)^2 + (\text{cosec } x)^2 - (\cot x)^2 \\ & = \sec^2 x - \tan^2 x + \text{cosec}^2 x - \cot^2 x \\ & = (1 + \tan^2 x) - \tan^2 x + \text{cosec}^2 x - \cot^2 x \phantom{0000000000} [\sec^2 A = 1 + \tan^2 A] \\ & = 1 + \text{cosec}^2 x - \cot^2 x \\ & = 1 + (1 + \cot^2 x) - \cot^2 x \phantom{0000000000000000000000} [\text{cosec}^2 A = 1 + \cot^2 A] \\ & = 1 + 1 \\ & = 2 \\ & = \text{R.H.S} \end{align}

\begin{align} \text{L.H.S} & = (\sin \alpha + \cos \beta)^2 - (\sin \alpha + \cos \beta)(\sin \alpha - \cos \beta) \\ & = (\sin \alpha)^2 + 2(\sin \alpha)(\cos \beta) + (\cos \beta)^2 - [ (\sin \alpha)^2 - (\cos \beta)^2 ] \\ & = \sin^2 \alpha + 2\sin \alpha \cos \beta + \cos^2 \beta - ( \sin^2 \alpha - \cos^2 \beta) \\ & = \sin^2 \alpha + 2\sin \alpha \cos \beta + \cos^2 \beta - \sin^2 \alpha + \cos^2 \beta \\ & = 2\sin \alpha \cos \beta + 2 \cos^2 \beta \\ & = 2\cos \beta ( \sin \alpha + \cos \beta) \\ & = \text{R.H.S} \end{align}

\begin{align} \sin x + \sin y & = a \\ \sin y & = a - \sin x \\ (\sin y)^2 & = (a - \sin x)^2 \\ \sin^2 y & = (a)^2 - 2(a)(\sin x) + (\sin x)^2 \\ \sin^2 y & = a^2 - 2a\sin x + \sin^2 x \phantom{000} \text{ --- (1)} \\ \\ \\ \cos x + \cos y & = a \\ \cos y & = a - \cos x \\ (\cos y)^2 & = (a - \cos x)^2 \\ \cos^2 y & = (a)^2 - 2(a)(\cos x) + (\cos x)^2 \\ \cos^2 y & = a^2 - 2a\cos x + \cos^2 x \phantom{000} \text{ --- (2)} \\ \\ \\ (1) & + (2), \\ \sin^2 y + \cos^2 y & = a^2 - 2a\sin x + \sin^2 x + a^2 - 2a\cos x + \cos^2 x \\ (\sin^2 y + \cos^2 y) & = 2a^2 - 2a\sin x - 2a\cos x + (\sin^2 x + \cos^2 x) \\ \\ \\ \text{Using the identity } & \sin^2 A + \cos^2 A = 1, \\ 1 & = 2a^2 - 2a\sin x - 2a \cos x + 1 \\ 0 & = 2a^2 - 2a\sin x - 2a\cos x \\ 2a\sin x + 2a\cos x & = 2a^2 \\ 2a(\sin x + \cos x) & = 2a^2 \\ \sin x + \cos x & = {2a^2 \over 2a} \\ \sin x + \cos x & = a \end{align}

(i)

\begin{align} k & = {1 + \sin x \over \cos x} \\ {k \over 1} & = {1 + \sin x \over \cos x} \\ (k)(\cos x) & = (1 + \sin x) \\ {\cos x \over 1 + \sin x} & = {1 \over k} \\ {\cos x (1 - \sin x) \over (1 + \sin x)(1 - \sin x)} & = {1 \over k} \\ {\cos x (1 - \sin x) \over (1)^2 - (\sin x)^2} & = {1 \over k} \\ {\cos x (1 - \sin x) \over 1 - \sin^2 x} & = {1 \over k} \\ \\ \\ \text{Since } \sin^2 A + \cos^2 A = 1 & \text{ and } \cos^2 A = 1 - \sin^2 A, \\ {\cos x(1 - \sin x) \over \cos^2 x} & = {1 \over k} \\ {1 - \sin x \over \cos x} & = {1 \over k} \text{ (Shown)} \end{align}

(ii)

\begin{align} k + {1 \over k} & = {1 + \sin x \over \cos x} + {1 - \sin x \over \cos x} \\ {k^2 \over k} + {1 \over k} & = {1 + \sin x + 1 - \sin x \over \cos x} \\ {k^2 + 1 \over k} & = {2 \over \cos x} \\ \cos x & = {2k \over k^2 + 1} \\ \\ \\ k & = {1 + \sin x \over \cos x} \\ k & = {1 + \sin x \over {2k \over k^2 + 1}} \phantom{00000} \text{[Use result for } \cos x] \\ k\left(2k \over k^2 + 1\right) & = 1 + \sin x \\ {2k^2 \over k^2 + 1} & = 1 + \sin x \\ \\ \sin x & = {2k^2 \over k^2 + 1} - 1 \\ & = {2k^2 \over k^2 + 1} - {k^2 + 1 \over k^2 + 1} \\ & = {2k^2 - (k^2 + 1) \over k^2 + 1} \\ & = {2k^2 - k^2 - 1 \over k^2 + 1} \\ & = {k^2 - 1 \over k^2 + 1} \end{align}

First part

\begin{align} \text{L.H.S} & = \sqrt{ 1 - \sin x \over 1 + \sin x } \\ & = \sqrt{ {1 - \sin x \over 1 + \sin x} \times {1 - \sin x \over 1 - \sin x} } \\ & = \sqrt{ (1 - \sin x)^2 \over (1 + \sin x)(1 - \sin x) } \\ & = \sqrt{ (1 - \sin x)^2 \over 1 - \sin^2 x } \\ & = \sqrt{ (1 - \sin x)^2 \over \cos^2 x } \phantom{000000} [\sin^2 A + \cos^2 A = 1 \rightarrow \cos^2 A = 1 - \sin^2 A] \\ & = {1 - \sin x \over \cos x} \\ & = {1 \over \cos x} - {\sin x \over \cos x} \\ & = \sec x - \tan x \\ & = \text{R.H.S} \end{align}

Second part

\begin{align} \sqrt{1 - \sin x \over 1 + \sin x} & > 0 \\ \\ \\ \sec x - \tan x & = {1 \over \cos x} - {\sin x \over \cos x} \\ & = {1 - \sin x \over \cos x} \\ \\ \text{If } 0^\circ < x < 90^\circ, & \phantom{.} 0 < \sin x < 1 \\ & \phantom{.} 0 < \cos x < 1 \\ & \phantom{.} {1 - \sin x \over \cos x} > 0 \\ \\ \text{Thus, identity } & \text{is valid for acute values of } x \end{align}

(i)

\begin{align} \text{If } 90^\circ < x < 180^\circ, & \phantom{.} 0 < \sin x < 1 \\ & \phantom{.} -1 < \cos x < 0 \\ & \phantom{.} {1 - \sin x \over \cos x} < 0 \\ \\ \\ \text{Since } \sqrt{1 - \sin x \over 1 + \sin x} & > 0, \\ \\ \therefore \sqrt{1 - \sin x \over 1 + \sin x} & = -(\sec x - \tan x) \\ \sqrt{1 - \sin x \over 1 + \sin x} & = \tan x - \sec x \end{align}

(ii)

\begin{align} \sqrt{1 - \sin x \over 1 + \sin x} & = \begin{cases} \sec x - \tan x, \text{ where } \cos x > 0 \\ \tan x - \sec x, \text{ where } \cos x < 0 \end{cases} \end{align}

\begin{align} & \phantom{0=} 4 \sin A \cos A + (\cos A - 5 \sin A)^2 + (3 \cos A + \sin A)^2 \\ & = 4 \sin A \cos A + \cos^2 A - 10 \sin A \cos A + 25 \sin^2 A + 9 \cos^2 A + 6 \sin A \cos A + \sin^2 A \\ & = 10 \cos^2 A + 26 \sin^2 A \\ & = 10 \cos^2 A + 26(1 - \cos^2 A) \phantom{000000} [\sin^2 A + \cos^2 A = 1 \rightarrow \sin^2 A = 1 - \cos^2 A] \\ & = 10 \cos^2 A + 26 - 26 \cos^2 A \\ & = -16 \cos^2 A + 26 \\ \\ \therefore p & = -16, q = 26 \end{align}