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