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