Trigonometry

Assertion: The value of each of the trigonometric ratios of an angle does not depend on the size of the triangle. It only depends on the angle.

Reason: In right as hypotenuse is the longest side.

Assertion: Sin 60^o = Cos 30^o

Reason: Sin 2θ = Sin θ where θ is an acute angle.

∵ sec θ is in the denominator

∴ The min. value of sec θ will return max. value for $\frac{1}{sec\theta }$.

But the min. value of sec θ is sec 0° = 1.

Hence, the max. value of $\frac{1}{sec{\theta }^{\circ }}$ = $\frac{1}{1}$ = 1

It means value of θ = 45°

Now, 2 tan θ + cos² θ = 2 tan 45° + cos² 45°

⇒ cos [90° – (x – 20)°] = cos (3x – 10)°

By comparing the coefficient

90° – x° + 20° = 3x° – 10° = 110° + 10° = 3x° + x°

120° = 4x°

$\Rightarrow \frac{{120}^{\circ }}{4}$ = 30°

⇒ sin2A = $\frac{1}{2}$ (1)² [∵ tan 45° = 1]

= sin2 A = $\frac{1}{2}$

⇒ sin A = $\frac{1}{\sqrt{2}}$

Hence, ∠A = 45°

b²x² + a²y² – a²b² = b²(acos θ)² + a²(b sin θ)² – a²b²

= a²b² cos²θ + a²b² sin² θ – a²b² = a²b² (sin² θ + cos² θ) – a²b²

= a²b² – a²b² = θ (∵ sin2 θ + cos² θ = 1)

tan A = cot B

⇒ tan A = tan (90° – B)

A = 90° – B

[∵ Both A and B are acute angles]

⇒ A + B = 90°

∴ ∠A + ∠B = 180° – ∠C = 90°

Now, sin² A + sin² B = sin² A + sin² (90° – A) = sin² A + cos² A = 1

sec 4 A = cosec (A – 20°)

⇒ cosec (90° – 4 A) = cosec (A – 20°)

∴ 90° – 4 A = A – 20°

⇒ 90° + 20° = A + 4 A

⇒ 110° = 5 A

∴ A = $\frac{110}{5}$ = 22°

Now, we know that

Now, we have, 15 cot A = 8

PR² = PO² + QR²

⇒ (13)2 = (12)² + QR²

⇒ 169 = 144 + QR²

⇒ QR² = 169 – 144 = 25

⇒ QR = 5 cm

Now, tan P = $\frac{QR}{PQ}=\frac{5}{12}$and $cotR=\frac{QR}{PQ}=\frac{5}{12}$

tan P – cot R = $\frac{5}{12}$ – $\frac{5}{12}$ = 0

⇒ (sin θ + cos θ)² = 3

⇒ sin² θ + cos² θ + 2 sin θ cos θ = 3

⇒ 2 sin cos θ = 2 (∵ sin² θ + cos² θ = 1)

⇒ sin θ. cos θ = 1 = sin² θ + cos² θ

⇒ 1 = tan θ + cot θ = 1

Therefore tan θ + cot θ = 1

= sin (90° – 65°). cos 65° + cos (90° – 65°). sin 65°

= cos 65° . cos 65° + sin 65°. sin 65°

= cos2 65° + sin2 65° = 1.

3 cos 68°. cosec 22° – $\frac{1}{2}$ tan 43°. tan 47°. tan 12°. tan 60°. tan 78°.

= 3 cos (90° – 22°). cosec 22° – $\frac{1}{2}$ . {tan 43° . tan (90° – 43°)}. {tan 12°. tan (90° – 12°). tan 60°}

= 3 sin 22°. cosec 22° – $\frac{1}{2}$ (tan 43° . cot 43°). (tan 12°. cot 12°). tan 60°

= 3 × 1 – × 1 × 1 × √3 $=3–\frac{3}{\sqrt{2}}=\frac{6-\sqrt{3}}{\surd 2}.$

We have a right-angled ∆PQR in which ∠Q = 90°.

Let QR = x cm

Therefore, PR = (25 – x) cm

By Pythagoras Theorem, we have

PR² = PQ² + QR²

(25 – x)² = 52 + x²

= (25 – x)² – x² = 25

(25 – x – x) (25 – x + x) = 25

(25 – 2x) 25 = 25

25 – 2x = 1

25 – 1 = 2x

= 24 = 2x

∴ x = 12 cm

Hence, QR = 12 cm

PR = (25 – x) cm = 25 – 12 = 13 cm

PQ = 5 cm

We have a right-angled ∆ABC in which ∠B = 90°.

and, $tanA=\frac{1}{\sqrt{3}}$

Now, tan A = $\frac{1}{\sqrt{3}}$ = BCAB

Let BC = k and AB = √3k

∴ By Pythagoras Theorem, we have

⇒ AC² = AB² + BC²

⇒ AC² = (√3k)² + (k)² = 3k² + k²

⇒ AC² = 4k²

Let us consider a right triangle ABC in which ∠B = 90°

Let AB = 4k and BC = 3k

∴ By Pythagoras Theorem

AC² = AB² + BC²

AC = (4k)² + (3k)² = 16k² + 9k²

AC² = 25k²

∴ AC = 5k

Let us consider a right-angled ∆ABC in which ∠B = 90°.

For ∠A we have

= 2 cosec² A tan² A = 2(1 + cot² A). tan² A

= 2 tan² A + 2 tan² A. cot² A (∵ tan A cot A = 1)

= 2 + 2 tan² A = 2(1 + tan² A) = 2 sec² A = RHS.

LHS = (sin θ + sec θ)² + (cos θ + cosec θ)²

= (1 + sec θ cosec θ)² = RHS.