In calculation exercises, the fundamental relationship gives the module and the quadrant gives the sign. In the identities, indicate which of the three methods you used.

Calculation of functions from one of them

73

Knowing that $\operatorname{sen} x = \dfrac{4}{5}$ and $\dfrac{\pi}{2} < x < \pi$, calculate the other circular functions of $x$.

See the resolution

As $\dfrac{\pi}{2} < x < \pi \Rightarrow \cos x < 0$, we have:

$$\cos x = -\sqrt{1 - \operatorname{sen}^2 x} = -\sqrt{1 - \frac{16}{25}} = -\sqrt{\frac{9}{25}} = -\frac{3}{5}$$$$\operatorname{tg} x = \frac{\operatorname{sen} x}{\cos x} = \frac{\frac{4}{5}}{-\frac{3}{5}} = -\frac{4}{3} \qquad \operatorname{cotg} x = \frac{\cos x}{\operatorname{sen} x} = \frac{-\frac{3}{5}}{\frac{4}{5}} = -\frac{3}{4}$$$$\sec x = \frac{1}{\cos x} = \frac{1}{-\frac{3}{5}} = -\frac{5}{3} \qquad \operatorname{cossec} x = \frac{1}{\operatorname{sen} x} = \frac{1}{\frac{4}{5}} = \frac{5}{4}.$$
74

Knowing that $\operatorname{cossec} x = -\dfrac{25}{24}$ and $\pi < x < \dfrac{3\pi}{2}$, calculate the other circular functions of $x$.

Ver resposta

$\operatorname{sen} x = -\dfrac{24}{25}$, $\cos x = -\dfrac{7}{25}$, $\operatorname{tg} x = \dfrac{24}{7}$, $\operatorname{cotg} x = \dfrac{7}{24}$, $\sec x = -\dfrac{25}{7}$

75

Knowing that $\operatorname{tg} x = \dfrac{12}{5}$ and $\pi < x < \dfrac{3\pi}{2}$, calculate the other circular functions of $x$.

See the resolution
$$\operatorname{cotg} x = \frac{1}{\operatorname{tg} x} = \frac{1}{\frac{12}{5}} = \frac{5}{12}.$$

As $\pi < x < \dfrac{3\pi}{2} \Rightarrow \sec x < 0$, we have:

$$\sec x = -\sqrt{1 + \operatorname{tg}^2 x} = -\sqrt{1 + \frac{144}{25}} = -\sqrt{\frac{169}{25}} = -\frac{13}{5} \qquad \cos x = \frac{1}{\sec x} = -\frac{5}{13}$$$$\operatorname{sen} x = \operatorname{tg} x \cdot \cos x = \left(\frac{12}{5}\right)\left(-\frac{5}{13}\right) = -\frac{12}{13} \qquad \operatorname{cossec} x = \frac{1}{\operatorname{sen} x} = -\frac{13}{12}.$$
76

Calculate $\cos x$ knowing that $\operatorname{cotg} x = \dfrac{2\sqrt{m}}{m - 1}$, with $m > 1$.

Ver resposta

$\cos x = \pm\dfrac{2\sqrt{m}}{m + 1}$

77

Calculate $\sec x$ knowing that $\operatorname{sen} x = \dfrac{2ab}{a^2 + b^2}$, with $a > b > 0$.

Ver resposta

$\sec x = \pm\dfrac{a^2 + b^2}{a^2 - b^2}$

78

Knowing that $\sec x = 3$, calculate the value of the expression $y = \operatorname{sen}^2 x + 2 \cdot \operatorname{tg}^2 x$.

See the resolution
$$\cos x = \frac{1}{\sec x} = \frac{1}{3} \qquad \operatorname{sen}^2 x = 1 - \cos^2 x = 1 - \frac{1}{9} = \frac{8}{9} \qquad \operatorname{tg}^2 x = \sec^2 x - 1 = 9 - 1 = 8$$

then

$$y = \operatorname{sen}^2 x + 2 \cdot \operatorname{tg}^2 x = \frac{8}{9} + 16 = \frac{152}{9}.$$
79

Being $\operatorname{sen} x = \dfrac{1}{3}$ and $0 < x < \dfrac{\pi}{2}$, calculate the value of

$$y = \frac{1}{\operatorname{cossec} x + \operatorname{cotg} x} + \frac{1}{\operatorname{cossec} x - \operatorname{cotg} x}.$$
Ver resposta

$y = 6$

80

Knowing that $\operatorname{cotg} x = \dfrac{24}{7}$ and $\pi < x < \dfrac{3\pi}{2}$, calculate the value of the expression

$$y = \frac{\operatorname{tg} x \cdot \cos x}{(1 + \cos x)(1 - \cos x)}.$$
See the resolution

1st solution. We calculate $\operatorname{tg} x$, $\cos x$ and finally $y$:

$$\operatorname{tg} x = \frac{1}{\operatorname{cotg} x} = \frac{7}{24} \qquad \cos^2 x = \frac{1}{1 + \operatorname{tg}^2 x} = \frac{1}{1 + \frac{49}{576}} = \frac{576}{625} \;\Rightarrow\; \cos x = -\frac{24}{25}$$$$y = \frac{\left(\frac{7}{24}\right)\left(-\frac{24}{25}\right)}{\left(1 - \frac{24}{25}\right)\left(1 + \frac{24}{25}\right)} = \frac{-\frac{7}{25}}{\frac{49}{625}} = -\frac{25}{7}.$$

2nd solution. We simplify $y$ and then calculate what is necessary:

$$y = \frac{\frac{\operatorname{sen} x}{\cos x} \cdot \cos x}{1 - \cos^2 x} = \frac{\operatorname{sen} x}{\operatorname{sen}^2 x} = \frac{1}{\operatorname{sen} x} = \operatorname{cossec} x = -\sqrt{1 + \operatorname{cotg}^2 x} = -\sqrt{1 + \frac{576}{49}} = -\frac{25}{7}.$$
81

Given that $\cos x = \dfrac{2}{5}$ and $\dfrac{3\pi}{2} < x < 2\pi$, obtain the value of $y = (1 + \operatorname{tg}^2 x)^2 + (1 - \operatorname{tg}^2 x)^2$.

Ver resposta

$y = \dfrac{457}{8}$

82

Calculate $\operatorname{sen} x$ and $\cos x$ knowing that $3 \cdot \cos x + \operatorname{sen} x = -1$.

See the resolution

We solve the system

$$\begin{cases} 3 \cdot \cos x + \operatorname{sen} x = -1 & (1) \\ \cos^2 x + \operatorname{sen}^2 x = 1 & (2) \end{cases}$$

From (1) comes $\operatorname{sen} x = -1 - 3 \cdot \cos x$ (1’). Replacing (1’) in (2):

$$\cos^2 x + (-1 - 3\cos x)^2 = 1 \;\Longrightarrow\; \cos^2 x + 1 + 6\cos x + 9\cos^2 x = 1 \;\Longrightarrow\; 10\cos^2 x + 6\cos x = 0,$$

then $\cos x = 0$ or $\cos x = -\dfrac{3}{5}$. Replacing each alternative in (1’):

$$\operatorname{sen} x = -1 - 3 \cdot 0 = -1 \qquad\text{ou}\qquad \operatorname{sen} x = -1 - 3\left(-\frac{3}{5}\right) = \frac{4}{5}.$$

There are two solutions: 1ª) $\cos x = 0$ and $\operatorname{sen} x = -1$; 2ª) $\cos x = -\dfrac{3}{5}$ and $\operatorname{sen} x = \dfrac{4}{5}$.

83

Calculate $\operatorname{sen} x$ and $\cos x$ knowing that $5 \cdot \sec x - 3 \cdot \operatorname{tg}^2 x = 1$.

Ver resposta

$\cos x = \dfrac{1}{2}$ e $\operatorname{sen} x = \pm\dfrac{\sqrt{3}}{2}$

84

Get $\operatorname{tg} x$ knowing that $\operatorname{sen}^2 x - 5 \cdot \operatorname{sen} x \cdot \cos x + \cos^2 x = 3$.

Ver resposta

$\operatorname{tg} x = -2$ or $\operatorname{tg} x = -\dfrac{1}{2}$

Conditions on parameters and deletion

85

Calculate $m$ so that you have $\operatorname{sen} x = 2m + 1$ and $\cos x = 4m + 1$.

See the resolution

As $\operatorname{sen}^2 x + \cos^2 x = 1$, it results:

$$(2m + 1)^2 + (4m + 1)^2 = 1 \;\Longrightarrow\; (4m^2 + 4m + 1) + (16m^2 + 8m + 1) = 1 \;\Longrightarrow\; 20m^2 + 12m + 1 = 0$$$$m = \frac{-12 \pm \sqrt{144 - 80}}{40} = \frac{-12 \pm 8}{40} \;\Longrightarrow\; m = -\frac{1}{2} \quad\text{ou}\quad m = -\frac{1}{10}.$$
86

Calculate $m$ so that you have $\operatorname{tg} x = m - 2$ and $\operatorname{cotg} x = \dfrac{m}{3}$.

Ver resposta

$m = 3$ or $m = -1$

87

Determine $a$ so that you have $\cos x = \dfrac{1}{a + 1}$ and $\operatorname{cossec} x = \dfrac{a + 1}{\sqrt{a + 2}}$.

Ver resposta

$a = 1$

88

Determine a relationship between $x$ and $y$, independent of $t$, knowing that $x = 3 \cdot \operatorname{sen} t$ and $y = 4 \cdot \cos t$.

See the resolution

As $\operatorname{sen}^2 t + \cos^2 t = 1$, it results:

$$\left(\frac{x}{3}\right)^2 + \left(\frac{y}{4}\right)^2 = 1 \;\Longrightarrow\; \frac{x^2}{9} + \frac{y^2}{16} = 1 \;\Longrightarrow\; 16x^2 + 9y^2 = 144.$$
89

Determine a relationship between $x$ and $y$, independent of $t$, knowing that $x = 5 \cdot \operatorname{tg} t$ and $y = 3 \cdot \operatorname{cossec} t$.

See the resolution

As $\operatorname{cossec}^2 t = \operatorname{cotg}^2 t + 1$ and $\operatorname{cotg} t = \dfrac{1}{\operatorname{tg} t}$, it results:

$$\left(\frac{y}{3}\right)^2 = \left(\frac{5}{x}\right)^2 + 1 \;\Longrightarrow\; \frac{y^2}{9} = \frac{25}{x^2} + 1 \;\Longrightarrow\; x^2y^2 = 225 + 9x^2 \;\Longrightarrow\; x^2y^2 - 9x^2 = 225.$$
90

If $\operatorname{sen} x + \cos x = a$ and $\operatorname{sen} x \cdot \cos x = b$, obtain a relationship between $a$ and $b$, independent of $x$.

Ver resposta

$a^2 - 2b = 1$

91

Given that $\operatorname{sen} x \cdot \cos x = m$, calculate the value of $y = \operatorname{sen}^4 x + \cos^4 x$ and $z = \operatorname{sen}^6 x + \cos^6 x$.

See the resolution

As $a^2 + b^2 \equiv (a + b)^2 - 2ab$, we have:

$$y = (\operatorname{sen}^2 x)^2 + (\cos^2 x)^2 = (\operatorname{sen}^2 x + \cos^2 x)^2 - 2\operatorname{sen}^2 x \cos^2 x = 1^2 - 2(\operatorname{sen} x \cos x)^2 = 1 - 2m^2.$$

As $a^3 + b^3 \equiv (a + b)(a^2 - ab + b^2)$, we have:

$$z = (\operatorname{sen}^2 x)^3 + (\cos^2 x)^3 = (\operatorname{sen}^2 x + \cos^2 x)(\operatorname{sen}^4 x - \operatorname{sen}^2 x \cos^2 x + \cos^4 x)$$$$= \operatorname{sen}^4 x + \cos^4 x - \operatorname{sen}^2 x \cos^2 x = y - (\operatorname{sen} x \cos x)^2 = 1 - 2m^2 - m^2 = 1 - 3m^2.$$
92

Knowing that $\operatorname{sen} x + \cos x = a$ ($a$ given), calculate $y = \operatorname{sen}^3 x + \cos^3 x$.

Ver resposta

$y = \dfrac{a}{2}(3 - a^2)$

Demonstration of identities

93

Prove that $(1 + \operatorname{cotg}^2 x)(1 - \cos^2 x) = 1$ for every real $x$, $x \ne k\pi$.

See the resolution
$$f(x) = (1 + \operatorname{cotg}^2 x)(1 - \cos^2 x) = \left(1 + \frac{\cos^2 x}{\operatorname{sen}^2 x}\right) \cdot \operatorname{sen}^2 x = \frac{\operatorname{sen}^2 x + \cos^2 x}{\operatorname{sen}^2 x} \cdot \operatorname{sen}^2 x = 1 = g(x).$$
94

Prove that $2 \cdot \sec x \cdot \operatorname{tg} x = \dfrac{1}{\operatorname{cossec} x - 1} + \dfrac{1}{\operatorname{cossec} x + 1}$ for all real $x$, $x \ne \dfrac{\pi}{2} + k\pi$ (and $x \ne k\pi$, so that $\operatorname{cossec} x$ exists).

See the resolution
$$g(x) = \frac{1}{\operatorname{cossec} x - 1} + \frac{1}{\operatorname{cossec} x + 1} = \frac{(\operatorname{cossec} x + 1) + (\operatorname{cossec} x - 1)}{(\operatorname{cossec} x - 1)(\operatorname{cossec} x + 1)} = \frac{2\operatorname{cossec} x}{\operatorname{cossec}^2 x - 1} = \frac{2\operatorname{cossec} x}{\operatorname{cotg}^2 x}$$$$= \frac{2}{\operatorname{sen} x} \cdot \frac{\operatorname{sen}^2 x}{\cos^2 x} = 2 \cdot \frac{1}{\cos x} \cdot \frac{\operatorname{sen} x}{\cos x} = 2 \cdot \sec x \cdot \operatorname{tg} x = f(x).$$
95

Prove that $(1 - \operatorname{tg} x)^2 + (1 - \operatorname{cotg} x)^2 = (\sec x - \operatorname{cossec} x)^2$ for every real $x$, $x \ne \dfrac{k\pi}{2}$.

See the resolution

We transform the two members separately to the same expression $h(x)$:

$$f(x) = \left(1 - \frac{\operatorname{sen} x}{\cos x}\right)^2 + \left(1 - \frac{\cos x}{\operatorname{sen} x}\right)^2 = \left(\frac{\cos x - \operatorname{sen} x}{\cos x}\right)^2 + \left(\frac{\operatorname{sen} x - \cos x}{\operatorname{sen} x}\right)^2$$$$= \frac{1 - 2\operatorname{sen} x \cos x}{\cos^2 x} + \frac{1 - 2\operatorname{sen} x \cos x}{\operatorname{sen}^2 x} = (1 - 2\operatorname{sen} x \cos x)\left(\frac{1}{\cos^2 x} + \frac{1}{\operatorname{sen}^2 x}\right) = \frac{1 - 2\operatorname{sen} x \cos x}{\cos^2 x \operatorname{sen}^2 x} = h(x)$$$$g(x) = \left(\frac{1}{\cos x} - \frac{1}{\operatorname{sen} x}\right)^2 = \left(\frac{\operatorname{sen} x - \cos x}{\cos x \operatorname{sen} x}\right)^2 = \frac{1 - 2\operatorname{sen} x \cos x}{\cos^2 x \operatorname{sen}^2 x} = h(x).$$
96

Prove that $\dfrac{1 - \cos x}{\operatorname{sen} x \cdot \cos x} + \operatorname{sen} x = \dfrac{1 - \cos x}{\operatorname{tg} x} + \operatorname{tg} x$ for every real $x$, $x \ne \dfrac{k\pi}{2}$.

See the resolution

We show that $f(x) - g(x) = 0$:

$$f(x) - g(x) = \frac{1 - \cos x}{\operatorname{sen} x \cos x} + \operatorname{sen} x - \frac{\cos x(1 - \cos x)}{\operatorname{sen} x} - \frac{\operatorname{sen} x}{\cos x}$$$$= \frac{1 - \cos x + \operatorname{sen}^2 x \cos x - \cos^2 x(1 - \cos x) - \operatorname{sen}^2 x}{\operatorname{sen} x \cos x}$$$$= \frac{1 - \cos x + (1 - \cos^2 x)\cos x - \cos^2 x(1 - \cos x) - (1 - \cos^2 x)}{\operatorname{sen} x \cos x}$$$$= \frac{1 - \cos x + \cos x - \cos^3 x - \cos^2 x + \cos^3 x - 1 + \cos^2 x}{\operatorname{sen} x \cos x} = 0.$$

In exercises 97 to 116, demonstrate the identities (in the values ​​of $x$ and $y$ in which all involved functions are defined). To shorten the demonstrations, we write $s = \operatorname{sen} x$ and $c = \cos x$.

97

$\cos^4 x + \operatorname{sen}^4 x + 2(\operatorname{sen} x \cdot \cos x)^2 = 1$

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The first member is the perfect square $(c^2 + s^2)^2 = 1^2 = 1$.

98

$\dfrac{\operatorname{sen} x}{\operatorname{cossec} x} + \dfrac{\cos x}{\sec x} = 1$

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$\dfrac{s}{1/s} + \dfrac{c}{1/c} = s^2 + c^2 = 1$.

99

$\operatorname{tg} x + \operatorname{cotg} x = \sec x \cdot \operatorname{cossec} x$

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$\dfrac{s}{c} + \dfrac{c}{s} = \dfrac{s^2 + c^2}{sc} = \dfrac{1}{sc} = \dfrac{1}{c} \cdot \dfrac{1}{s} = \sec x \cdot \operatorname{cossec} x$.

100

$(\operatorname{tg} x + \operatorname{cotg} x)(\sec x - \cos x)(\operatorname{cossec} x - \operatorname{sen} x) = 1$

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For the previous year, $\operatorname{tg} x + \operatorname{cotg} x = \dfrac{1}{sc}$. Also, $\sec x - c = \dfrac{1 - c^2}{c} = \dfrac{s^2}{c}$ and $\operatorname{cossec} x - s = \dfrac{1 - s^2}{s} = \dfrac{c^2}{s}$. The product is $\dfrac{1}{sc} \cdot \dfrac{s^2}{c} \cdot \dfrac{c^2}{s} = 1$.

101

$\sec^2 x + \operatorname{cossec}^2 x = \sec^2 x \cdot \operatorname{cossec}^2 x$

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$\dfrac{1}{c^2} + \dfrac{1}{s^2} = \dfrac{s^2 + c^2}{c^2 s^2} = \dfrac{1}{c^2} \cdot \dfrac{1}{s^2}$.

102

$\dfrac{\operatorname{cotg}^2 x}{1 + \operatorname{cotg}^2 x} = \cos^2 x$

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$1 + \operatorname{cotg}^2 x = \operatorname{cossec}^2 x = \dfrac{1}{s^2}$, so the first member is $\dfrac{c^2}{s^2} \cdot s^2 = c^2$.

103

$\dfrac{\operatorname{sen}^3 x - \cos^3 x}{\operatorname{sen} x - \cos x} = 1 + \operatorname{sen} x \cdot \cos x$

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Like $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$, the first member is $s^2 + sc + c^2 = 1 + sc$.

104

$\operatorname{cossec}^2 x + \operatorname{tg}^2 x = \sec^2 x + \operatorname{cotg}^2 x$

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$\operatorname{cossec}^2 x + \operatorname{tg}^2 x = (1 + \operatorname{cotg}^2 x) + (\sec^2 x - 1) = \sec^2 x + \operatorname{cotg}^2 x$.

105

$2(\operatorname{sen} x + \operatorname{tg} x)(\cos x + \operatorname{cotg} x) = (1 + \operatorname{sen} x + \cos x)^2$

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$$2\left(s + \frac{s}{c}\right)\left(c + \frac{c}{s}\right) = 2 \cdot \frac{s(c + 1)}{c} \cdot \frac{c(s + 1)}{s} = 2(1 + c)(1 + s) = 2 + 2s + 2c + 2sc$$$$(1 + s + c)^2 = 1 + s^2 + c^2 + 2s + 2c + 2sc = 2 + 2s + 2c + 2sc.$$
106

$(1 + \operatorname{cotg} x)^2 + (1 - \operatorname{cotg} x)^2 = 2 \cdot \operatorname{cossec}^2 x$

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Developing the squares, the terms $\pm 2\operatorname{cotg} x$ cancel each other out: $2 + 2\operatorname{cotg}^2 x = 2(1 + \operatorname{cotg}^2 x) = 2\operatorname{cossec}^2 x$.

Note: in the printed edition, the first parenthesis appears without the exponent 2; without it, equality does not hold (for example, for $x = \frac{\pi}{4}$, the first member gives 2 and the second, 4).

107

$\dfrac{1 - 2\cos^2 x + \cos^4 x}{1 - 2\operatorname{sen}^2 x + \operatorname{sen}^4 x} = \operatorname{tg}^4 x$

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Numerator and denominator are perfect squares: $\dfrac{(1 - c^2)^2}{(1 - s^2)^2} = \dfrac{(s^2)^2}{(c^2)^2} = \dfrac{s^4}{c^4} = \operatorname{tg}^4 x$.

108

$(\operatorname{cotg} x - \cos x)^2 + (1 - \operatorname{sen} x)^2 = (1 - \operatorname{cossec} x)^2$

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$\operatorname{cotg} x - c = \dfrac{c - sc}{s} = \dfrac{c(1 - s)}{s}$. Then

$$\frac{c^2(1 - s)^2}{s^2} + (1 - s)^2 = (1 - s)^2 \cdot \frac{c^2 + s^2}{s^2} = \left(\frac{1 - s}{s}\right)^2 = (\operatorname{cossec} x - 1)^2 = (1 - \operatorname{cossec} x)^2.$$
109

$\dfrac{\cos x + \cos y}{\operatorname{sen} x - \operatorname{sen} y} = \dfrac{\operatorname{sen} x + \operatorname{sen} y}{\cos y - \cos x}$

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Two fractions are equal when the “cross products” are equal (non-zero denominators):

$$(\cos x + \cos y)(\cos y - \cos x) = \cos^2 y - \cos^2 x \qquad (\operatorname{sen} x - \operatorname{sen} y)(\operatorname{sen} x + \operatorname{sen} y) = \operatorname{sen}^2 x - \operatorname{sen}^2 y = (1 - \cos^2 x) - (1 - \cos^2 y) = \cos^2 y - \cos^2 x.$$
110

$\dfrac{\cos x + \operatorname{cotg} x}{\operatorname{tg} x + \sec x} = \cos x \cdot \operatorname{cotg} x$

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$$\frac{c + \frac{c}{s}}{\frac{s}{c} + \frac{1}{c}} = \frac{\frac{c(s + 1)}{s}}{\frac{s + 1}{c}} = \frac{c^2}{s} = c \cdot \frac{c}{s} = \cos x \cdot \operatorname{cotg} x.$$
111

$\dfrac{\operatorname{sen}^2 x - \cos^2 y}{\cos^2 x \cdot \cos^2 y} + 1 = \operatorname{tg}^2 x \cdot \operatorname{tg}^2 y$

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$$\frac{\operatorname{sen}^2 x - \cos^2 y + \cos^2 x \cos^2 y}{\cos^2 x \cos^2 y} = \frac{\operatorname{sen}^2 x - \cos^2 y(1 - \cos^2 x)}{\cos^2 x \cos^2 y} = \frac{\operatorname{sen}^2 x(1 - \cos^2 y)}{\cos^2 x \cos^2 y} = \frac{\operatorname{sen}^2 x \operatorname{sen}^2 y}{\cos^2 x \cos^2 y}.$$
112

$\dfrac{1 - \cos x}{1 + \cos x} = (\operatorname{cossec} x - \operatorname{cotg} x)^2$

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$$(\operatorname{cossec} x - \operatorname{cotg} x)^2 = \left(\frac{1 - c}{s}\right)^2 = \frac{(1 - c)^2}{1 - c^2} = \frac{(1 - c)^2}{(1 - c)(1 + c)} = \frac{1 - c}{1 + c}.$$
113

$\dfrac{\operatorname{cotg} x + \operatorname{cotg} y}{\operatorname{tg} x + \operatorname{tg} y} = \operatorname{cotg} x \cdot \operatorname{cotg} y$

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$$\frac{\frac{1}{\operatorname{tg} x} + \frac{1}{\operatorname{tg} y}}{\operatorname{tg} x + \operatorname{tg} y} = \frac{\frac{\operatorname{tg} x + \operatorname{tg} y}{\operatorname{tg} x \operatorname{tg} y}}{\operatorname{tg} x + \operatorname{tg} y} = \frac{1}{\operatorname{tg} x \operatorname{tg} y} = \operatorname{cotg} x \operatorname{cotg} y.$$
114

$(\sec x \cdot \sec y + \operatorname{tg} x \cdot \operatorname{tg} y)^2 = 1 + (\sec x \cdot \operatorname{tg} y + \sec y \cdot \operatorname{tg} x)^2$

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We use the 3rd method: the difference between the squares is a product of sum and difference.

$$(\sec x \sec y + \operatorname{tg} x \operatorname{tg} y)^2 - (\sec x \operatorname{tg} y + \sec y \operatorname{tg} x)^2$$$$= (\sec x - \operatorname{tg} x)(\sec y - \operatorname{tg} y) \cdot (\sec x + \operatorname{tg} x)(\sec y + \operatorname{tg} y) = (\sec^2 x - \operatorname{tg}^2 x)(\sec^2 y - \operatorname{tg}^2 y) = 1 \cdot 1 = 1.$$
115

$\sec x - \operatorname{tg} x = \dfrac{1}{\sec x + \operatorname{tg} x}$

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$(\sec x - \operatorname{tg} x)(\sec x + \operatorname{tg} x) = \sec^2 x - \operatorname{tg}^2 x = 1$; just divide by $\sec x + \operatorname{tg} x$ (which does not cancel, as the product is worth 1).

116

$\operatorname{cossec}^6 x - \operatorname{cotg}^6 x = 1 + 3 \cdot \operatorname{cotg}^2 x \cdot \operatorname{cossec}^2 x$

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Let $a = \operatorname{cossec}^2 x$ and $b = \operatorname{cotg}^2 x$, with $a - b = 1$. Then

$$a^3 - b^3 = (a - b)(a^2 + ab + b^2) = 1 \cdot \big[(a - b)^2 + 3ab\big] = 1 + 3ab = 1 + 3\operatorname{cotg}^2 x \operatorname{cossec}^2 x.$$