Solution:

Here,

$$\angle$$ABC = $$\theta$$, angle of reference.

$$\angle$$ABC = 90°

Hence,

AC = h, AB = p and BC = b

Solution:

Here,

$$\angle$$QPR = α, angle of reference.

$$\angle$$PQR = 90°

Hence,

RP = h, RQ = p and PQ = b

Solution:

Here,

$$\angle$$EFG = β, angle of reference.

$$\angle$$GEF 90°

Hence,

FG = h, EG = p and EF = b.

Solution:

Here,

$$\angle$$ABC = 90°

Hence,

or, $$\theta$$ + 35° = 90°

or, $$\theta$$ = 90°− 35°

∴ $$\theta$$ = 55°

Solution:

$$\angle$$XYZ = 90°

Here,

or, β + 56° = 90°

or, β = 90°− 56°

∴ β = 34°

Solution:

$$\angle$$QPR = 90°

Here,

or, α + 28 = 90°

or, α = 90°− 28°

∴ α = 52°

Solution,

Here,

AB = P = 3cm

BC = b = 4cm

AC = h = ?

From the right angled ΔABC we have,

or, h = $$\sqrt{p^2+b^2}$$

or, h = $$\sqrt{3^2+4^2}$$

or, h = $$\sqrt{9+16}$$

or, h = $$\sqrt{25}$$

∴ h = 5cm

Solution:

Here,

PQ = h = 13cm

RQ = p = 12cm

PR = b = ?

From right angled ΔPQR we have,

or, b = $$\sqrt{h^2−p^2}$$

or, b = $$\sqrt{13^2−12^2}$$

or, b = $$\sqrt{169−12}$$

or, b = $$\sqrt{25}$$

∴ b = 5cm

Solution:

Here,

YZ = h = 10cm

XY = b = 6cm

XZ = p = ?

From right angled ΔXYZ we have,

or, p = $$\sqrt{h^2−b^2}$$

or, p = $$\sqrt{10^2−6^2}$$

or, p = $$\sqrt{100−36}$$

or, p = $$\sqrt{64}$$

∴ p = 8cm

Solution:

The three sides of a triangle are 5cm, 12cm and 13cm

Here,

or, 132 = 52 + 122

or, 169 = 25 + 144

∴ 169 = 169

i.e. h2 = p2+ b2

Hence, the triangle is right angle triangle.

Use the Pythagorean theorem, with a = 12 and b = 5.

or, c2= a2+ b2[Pythagorean theorem]

or, c2= 122+ 52 [Putting value of a & b]

or, c2= 144 + 25

or, c2= 169

or, $$\sqrt{c^2}$$ = $$\sqrt{169}$$ [Squaring on both sides]

$$\therefore$$ c =13

Hence, The length of the hypotenuse is 13 metres.

Solution:

Use the Pythagorean theorem, with a = 3 and b = 4.

or, c2= a2+ b2[Pythagorean theorem]

or, c2= 32+ 42[Putting value of a & b]

or, c2= 9 + 16

or, c2= 25

or, $$\sqrt{c^2}$$ = $$\sqrt{25}$$ [Squaring on both side]

$$\therefore$$ c = 5

Hence, The length of the hypotenuse is 5 metres.

Solution:

L.H.S = $$\sqrt{\frac{secβ + 1}{secβ− 1}}$$
= $$\sqrt{\frac{secβ + 1}{secβ− 1}× \frac{secβ + 1}{secβ + 1}}$$
= $$\sqrt{\frac{(secβ + 1)^2}{secβ^2 − 1}}$$
= $$\sqrt{\frac{(secβ + 1)^2}{tan^2β}}$$
= $$\sqrt{(\frac{sectanβ + 1}{tanβ})^2}$$
= $$\frac{secβ + 1}{tanβ}$$
= $$\frac{\frac{1}{cosβ} + 1}{\frac{sinβ}{cosβ}}$$
= $$\frac{\frac{1 + cosβ}{cosβ}}{\frac{sinβ}{cosβ}}$$
= $$\frac{1 + cosβ}{cosβ}$$ × $$\frac{cosβ}{sinβ}$$
= $$\frac{1 + cosβ}{sinβ}$$
= R.H.S proved

Solution:

L.H.S = $$\frac{tanθ + tanβ}{cotθ + cotβ}$$
= $$\frac{\frac{sinθ}{cosθ} + \frac{sinβ}{cosβ}}{\frac{cosθ}{sinθ} + \frac{cosβ}{sinβ}}$$
= $$\frac{\frac{sinθ . cosβ + sinβ . cosθ}{cosθ . cosβ}}{\frac{cosθ . sinβ + cosβ . sinθ}{sinθ . sinβ}}$$
=$$\frac{sinθ . cosβ + sinβ . cosθ}{cosθ . cosβ}$$ × $$\frac{sinθ . sinβ }{cosθ . sinβ + cosβ . sinθ }$$
= $$\frac{sinθ . sinβ}{cosθ . cosβ}$$
= $$\frac{sinθ}{cosθ}$$ .$$\frac{sinβ}{cosβ}$$
= tanθ . tanβ
= R.H.S proved

Solution:

l.H.S = $$\frac{cosα + sinα}{cosα− sinα}$$ [Dividing numerator and denominator by cosα]
= $$\frac{\frac{cosα + sinα}{cosα}}{\frac{cosα− sinα}{cosα}}$$
= $$\frac{\frac{cosα}{cosα} + \frac{sinα}{cosα}}{\frac{cosα}{cosα} − \frac{sinα}{cosα}}$$
= $$\frac{1 + tanα}{1− tanα}$$ (Middle Term)

Middle Term = $$\frac{1 + tanα}{1− tanα}$$
= $$\frac{1 + \frac{1}{cotα}}{1 −\frac{1}{cotα}}$$
= $$\frac{cotα + 1}{cotα}$$× $$\frac{cotα }{cotα− 1}$$
= $$\frac{cotα + 1}{cotα− 1}$$ (R.H.S)
$$\therefore$$ L.H.S = M.T = R.H.S proved

Solution:

L.H.S = $$\frac{1}{1 + cosθ}$$ + $$\frac{1}{1− cosθ}$$
= $$\frac{1(1− cosθ) + 1(1 + cosθ)}{(1 + cosθ)(1− cosθ)}$$
= $$\frac{2}{1− cos^2θ}$$
= $$\frac{2}{sin^2θ}$$
= $$\frac{2}{\frac{1}{cosec^2θ}}$$
= 2cosec2θ
= R.H.S proved

Solution:

L.H.S = $$\frac{tanθ}{1− cotθ}$$ + $$\frac{cotθ}{1− tanθ}$$
= $$\frac{\frac{sinθ}{cosθ}}{1− \frac{cosθ}{sinθ}}$$ +$$\frac{\frac{cosθ}{sinθ}}{1− \frac{sinθ}{cosθ}}$$
= $$\frac{\frac{sinθ}{cosθ}}{\frac{sinθ−cosθ}{sinθ}}$$+$$\frac{\frac{cosθ}{sinθ}}{\frac{cosθ−sinθ}{cosθ}}$$
= $$\frac{sinθ}{cosθ}$$× $$\frac{sinθ}{(sinθ−cosθ)}$$ + $$\frac{cosθ}{sinθ}$$×$$\frac{cosθ}{(cosθ−sinθ)}$$
= $$\frac{sin^2θ}{cosθ(sinθ−cosθ)}$$ + $$\frac{cos^2θ}{sinθ(cosθ−sinθ)}$$
= $$\frac{sin^2θ}{cosθ(sinθ−cosθ)}$$−$$\frac{cos^2θ}{sinθ(sinθ−cosθ)}$$
= $$\frac{sin^2θ× sinθ−cos^2θ×cosθ}{sinθ . cosθ(sinθ−cosθ)}$$
= $$\frac{sin^3θ−cos^3θ}{sinθ . cosθ(sinθ−cosθ)}$$
= $$\frac{(sinθ−cosθ)(sin^2θ +sinθ . cosθ +cos^2θ)}{sinθ . cosθ(sinθ−cosθ)}$$
= $$\frac{(sin^2θ +sinθ . cosθ +cos^2θ)}{sinθ . cosθ}$$
= $$\frac{sin^2θ +cos^2θ +sinθ . cosθ}{sinθ . cosθ}$$
= $$\frac{1 +sinθ . cosθ}{sinθ . cosθ}$$
= $$\frac{1}{sinθ . cosθ}$$ +$$\frac{sinθ . cosθ}{sinθ . cosθ}$$
= $$\frac{1}{sinθ . cosθ}$$ + 1
= $$\frac{1}{sinθ}$$× $$\frac{1}{cosθ}$$ + 1
= secθ . cosecθ + 1
= R.H.S proved