Solution:

From the given figure,

Hypotenuse (h) = 13cm

Perpendicular (p) = 5cm

Base (b) = 12cm

Now,

$$\therefore$$ sinα = $$\frac{p}{h}$$ = $$\frac{5}{13}$$

$$\therefore$$ cosα = $$\frac{b}{h}$$ = $$\frac{12}{13}$$

$$\therefore$$ tanα = $$\frac{p}{b}$$ = $$\frac{5}{12}$$

Solution:

From the given figure,

Hypotenuse (h) = 10cm

Perpendicular (p) = 6cm

Base (b) = 8cm

Now,

$$\therefore$$ cosecθ = $$\frac{h}{p}$$ = $$\frac{10}{6}$$ = $$\frac{5}{3}$$

$$\therefore$$ secθ = $$\frac{h}{b}$$ = $$\frac{10}{8}$$ = $$\frac{5}{4}$$

$$\therefore$$ cotθ = $$\frac{b}{p}$$ = $$\frac{8}{6}$$ = $$\frac{4}{3}$$

Solution:

= 18sinθ − 7sinθ

= (18 − 7)sinθ

= 11sinθ

Solution:

L.H.S. = $$\frac{sinθ.cosecθ}{secθ}$$

= sinθ cosecθ÷ secθ

= sinθ× $$\frac{1}{sinθ}$$÷ $$\frac{1}{cosθ}$$

= sinθ× $$\frac{1}{sinθ}$$× cosθ

= cosθ

= R.H.S. proved.

Solution:

L.H.S. = secθ $$\sqrt{1− cos^2θ}$$

= $$\frac{1}{cosθ}$$ . sinθ

= $$\frac{sinθ}{cosθ}$$

= tanθ

= R.H.S. proved.

Solution:

L.H.S. = sin2θ + cos2θ.tan2θ

= sin2θ + cos2θ $$\frac{sin^2θ}{cos^2θ}$$

= sin2θ + sin2θ

= 2sin2θ

= R.H.S. proved

Solution:

L.H.S. = (sinA + cosA)2 + (sinA − cosA)2

= sin2A + 2sinA.cosA + cos2A + sin2A − 2sinA.cosA + cos2A

= (sin2A + cos2A) + (sin2A + cos2A)

= 1 + 1

= 2

= R.H.S. proved

Solution:

L.H.S. = $$\frac{1}{tanA + cotA}$$

= $$\frac {1}{\frac {sinA}{cosA} + \frac {cosA}{sinA}}$$

= $$\frac{1}{\frac{sin^2A + cos^2A}{sinA.cosA}}$$

= $$\frac{1}{\frac{1}{sinA.cosA}}$$

= sinA.cosA

= R.H.S. proved

Solution:

L.H.S. = $$\frac{sin^4θ− cos^4θ}{sinθ + cosθ}$$

= $$\frac{(sin^2θ)^2 − (cos^2θ)^2}{sinθ + cosθ}$$

= $$\frac{(sin^2θ + cos^2θ) (sin^2θ− cos^2θ)}{sinθ + cosθ}$$

= $$\frac{1× (sinθ + cosθ) (sinθ− cosθ)}{sinθ + cosθ}$$

= sinθ− cosθ

= R.H.S. proved

Solution:

L.H.S. = $$\frac{cosA}{1 + sinA} + \frac{cosA}{1− sinA}$$

= $$\frac{cosA (1− sinA) + cosA (1 + sinA)}{(1 + sinA) (1− sinA)}$$

= $$\frac{cosA− sinA.cosA + sinA.cosA}{1− sin^2A}$$

= $$\frac{2cosA}{cos^2A}$$

= $$\frac{2}{cosA}$$

= 2secA

= R.H.S. proved

Solution:

L.H.S.
= $$\frac{1− sin^4}{cos4A}$$

= $$\frac{1− sin^4A}{cos^4A}$$ − $$\frac{sin^4A}{cos^4A}$$

= sec4A − tan4A

= (sec2A)2− (tan2A)2

= (sec2A− tan2A) (sec2A + tan2A)

= (1 + tan2A + tan2A)

= 1 + 2tan2A

= R.H.S. proved

Solution:

L.H.S
= (xcosθ + ysinθ)2+ (xsinθ− yccosθ)2

= (xcosθ)2+ 2xysinθ.cosθ + (ysinθ)2+ (xsinθ)2− 2xysinθ.cosθ + (ycosθ)2

= x2cos2θ + y2sin2θ + x2sin2θ+ y2cos2θ

= x2(cos2θ + sin2θ) + y2(sin2θ + cos2θ)

= (x2+ y2) (cos2θ + sin2θ)

= (x2+ y2) $$\times$$ 1 [$$\because$$ (cos2θ + sin2θ) =1]

= (x2+ y2) R.H.S. proved

Solution:

L.H.S. =$$\sqrt\frac{1 + cosθ}{1 – cosθ}$$

= $$\sqrt{\frac{1 + cosθ}{1− cosθ}× \frac{1 + cosθ}{1− cosθ}}$$

= $$\sqrt\frac{(1 + cosθ)^2}{1− cos^2θ}$$

= $$\sqrt\frac{(1 + cosθ)^2}{sin^2θ}$$

= $$\frac{1 + cosθ}{sinθ}$$

= $$\frac{1}{sinθ} + \frac{cosθ}{sinθ}$$

= cosecθ + cotθ

= R.H.S. proved

Solution:

Let, ABC be a right angled triangle where ∠B = 90° and ∠C = θ = k

Let, sinθ = k

Then,

or, sinθ = $$\frac{k}{1}$$ = $$\frac{p}{h}$$

∴ p = k, h = 1

Now, from right angled ΔABC,

or, b = $$\sqrt{h^2 − p^2}$$

or, b = $$\sqrt{1 − k^2}$$

or, b = $$\sqrt{1− sin^2θ}$$

Then,

cosecθ = $$\frac{h}{p}$$ = $$\frac{1}{k}$$ = $$\frac{1}{sinθ}$$

cosθ = $$\frac{b}{h}$$ = $$\sqrt\frac{1− sin^2θ}{1}$$ = $$\sqrt{1−sin^2θ}$$

secθ = $$\frac{h}{b}$$ = $$\frac{1}{\sqrt{1− k^2}}$$

tanθ = $$\frac{p}{b}$$ = $$\frac{k}{\sqrt{1− k^2}}$$ = $$\frac{sinθ}{\sqrt{1− sin^2θ}}$$

cotθ = $$\frac{b}{p}$$ = $$\frac{\sqrt{1− sin^2θ}}{sinθ}$$

Solutions:

Here,

cosα = $$\frac{5}{13}$$ = $$\frac{b}{h}$$

Then,

b = 5, h = 13, p= ?

By using Pythagoras theorem, we have,

or, p2= h2− b2

or, p = $$\sqrt{h^2 − b^2}$$

or, p = $$\sqrt{13^2 − 5^2}$$

or, p = $$\sqrt{169 − 25}$$

or, p = $$\sqrt{144}$$

∴ p = 12

Now,

secα = $$\frac{h}{b}$$ = $$\frac{13}{5}$$

sinα = $$\frac{p}{h}$$ = $$\frac{12}{13}$$

cosecα = $$\frac{h}{p}$$ = $$\frac{13}{12}$$

tanα = $$\frac{p}{b}$$ = $$\frac{12}{5}$$

cotα = $$\frac{b}{p}$$ = $$\frac{5}{12}$$

Solution:

Here,

or, sinθ − cosθ = 0

or, sinθ = cosθ

or, tanθ = 1

We have,

tanθ = $$\frac{p}{b}$$ = $$\frac{1}{1}$$

Then,

p = 1, b = 1

By Pythagoras theorem,

or, h2 = p2 + b2

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

or, h = $$\sqrt{1^2 +1^2}$$

or, h = $$\sqrt{1 + 1}$$

or, h = $$\sqrt{2}$$

∴ h = $$\sqrt{2}$$

Now,

cosecθ = $$\frac{h}{p}$$ = $$\frac{\sqrt{2}}{1}$$ = $$\sqrt{2}$$

sinθ = $$\frac{p}{h}$$ = $$\frac{1}{\sqrt{2}}$$

Solution:

Here,

or, sinθ = $$\frac{m}{n}$$ = $$\frac{p}{h}$$

Then,

p = m, h = n

We have,

or, b2 = h2− p2

or, b2 = n2− m2

or, b = $$\sqrt{n^2 − m^2}$$

Now,

or, tanθ = $$\frac{p}{b}$$

or, tanθ = $$\frac{m}{\sqrt{n^2− m^2}}$$

or, $$\sqrt{n^2− m^2}$$ tanθ = m

$$\therefore$$ L.H.S = R.H.S proved

Solution:

Here,

or, 3cotθ = 4

We have,

or, cotθ = $$\frac{b}{p}$$ = $$\frac{4}{3}$$

Then,

b = 4, p = 3, h = ?

By Pythagoras theorem,

or, h2 = p2 + b2

or, h2 = 32 + 42

or, h2 = 9 + 16

or, h2 = $$\sqrt{25}$$

∴ h = 5

Now,

= $$\frac{3cosθ + 2sinθ}{3cosθ− 2sinθ}$$

= {3 $$(\frac{b}{h}) + 2 (\frac{p}{h})$$}÷ {3 $$(\frac{b}{h})− 2 (\frac{p}{h})$$}

= (3× $$\frac{4}{5}$$ + 2× $$\frac{3}{5}$$)÷ (3× $$\frac{4}{5}$$− 2× $$\frac{3}{5}$$)

= ($$\frac{12}{5} + \frac{6}{5}$$)÷ ($$\frac{12}{5}− \frac{6}{5}$$)

= $$\frac{18}{5}$$÷ $$\frac{6}{5}$$

= $$\frac{18}{5}$$× $$\frac{5}{6}$$

= 3