Trigonometrical Identities — Chapter Test

If θ is an acute angle and cosec θ = $\sqrt{5}$ find the value of cot θ - cos θ.

Answer

sin θ = $\dfrac{1}{\text{cosec θ}} = \dfrac{1}{\sqrt{5}}$,

cos² θ = 1 - sin² θ = 1 - $\Big(\dfrac{1}{\sqrt{5}}\Big)^2 = 1 - \dfrac{1}{5} = \dfrac{4}{5}.$

cos θ = $\sqrt{\dfrac{4}{5}}$ = $\dfrac{2}{\sqrt{5}}$.

cot θ = $\dfrac{\text{cos θ}}{\text{sin θ}} = \dfrac{\dfrac{2}{\sqrt{5}}}{\dfrac{1}{\sqrt{5}}}$ = 2.

$\text{cot θ - cos θ} = 2 - \dfrac{2}{\sqrt{5}} \\[1em] = 2\Big(1 - \dfrac{1}{\sqrt{5}}\Big) \\[1em] = \dfrac{2(\sqrt{5} - 1)}{\sqrt{5}}.$

Hence, the value of cot θ - cos θ = $\dfrac{2(\sqrt{5} - 1)}{\sqrt{5}}$.

If θ is an acute angle and tan θ = $\dfrac{8}{15}$, find the value of sec θ + cosec θ.

Answer

sec² θ = 1 + tan² θ

sec² θ = 1 + $\Big(\dfrac{8}{15}\Big)^2$

sec² θ = 1 + $\dfrac{64}{225} = \dfrac{225 + 64}{225} = \dfrac{289}{225}$.

sec θ = $\sqrt{\dfrac{289}{225}} = \dfrac{17}{15}$.

cot θ = $\dfrac{1}{\text{tan θ}} = \dfrac{1}{\dfrac{8}{15}} = \dfrac{15}{8}.$

cosec² θ = 1 + cot² θ

cosec² θ = 1 + $\Big(\dfrac{15}{8}\Big)^2$

cosec² θ = 1 + $\dfrac{225}{64} = \dfrac{64 + 225}{64} = \dfrac{289}{64}$.

cosec θ = $\sqrt{\dfrac{289}{64}} = \dfrac{17}{8}$.

$\text{sec θ + cosec θ} = \dfrac{17}{15} + \dfrac{17}{8} \\[1em] = \dfrac{17 \times 8 + 17 \times 15}{120} \\[1em] = \dfrac{136 + 255}{120} \\[1em] = \dfrac{391}{120} \\[1em] = 3\dfrac{31}{120}.$

Hence, the value of expression sec θ + cosec θ = $3\dfrac{31}{120}.$

Evaluate the following :

$2 \times \Big(\dfrac{\text{cos}^2 20° + \text{cos}^2 70°}{\text{sin}^2 25° + \text{sin}^2 65°}\Big)$ - tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°

Answer

Since, angles are acute in the equation,

∴ cos(90° - θ) = sin θ, sin(90° - θ) = cos θ and tan(90° - θ) = cot θ.

Using above equations in

2 x $\Big(\dfrac{\text{cos}^2 20° + \text{cos}^2 70°}{\text{sin}^2 25° + \text{sin}^2 65°}\Big)$ - tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°

= 2 x $\Big(\dfrac{\text{cos}^2 20° + \text{cos}^2 (90 - 20)°}{\text{sin}^2 25° + \text{sin}^2 (90 - 25)°}\Big)$ - tan 45° + tan 13° tan 23° tan 30° tan (90 - 23)° tan (90 - 13)°

= 2 x $\Big(\dfrac{\text{cos}^2 20° + \text{sin}^2 20°}{\text{sin}^2 25° + \text{cos}^2 25°}\Big)$ - tan 45° + tan 13° tan 23° tan 30° cot 23° cot 13°

= 2 x 1 - 1 + tan 13° cot 13° tan 23° cot 23° tan 30°

= 1 + 1 x $\dfrac{1}{\sqrt{3}}$

= $\dfrac{\sqrt{3} + 1}{\sqrt{3}}$

= $\dfrac{(\sqrt{3} + 1)\sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

= $\dfrac{3 + \sqrt{3}}{\sqrt{3}}$.

Hence, the value of the expression is $\dfrac{3 + \sqrt{3}}{\sqrt{3}}$.

Evaluate the following :

$\dfrac{\text{sin}^2 22° + \text{sin}^2 68°}{\text{cos}^2 22° + \text{cos}^2 68°}$ + sin² 63° + cos 63° sin 27°

Answer

Since, angles are acute in the equation,

∴ cos(90° - θ) = sin θ, sin(90° - θ) = cos θ.

Using above equations in

$\dfrac{\text{sin}^2 22° + \text{sin}^2 68°}{\text{cos}^2 22° + \text{cos}^2 68°}$ + sin² 63° + cos 63° sin 27°

= $\dfrac{\text{sin}^2 22° + \text{sin}^2 (90 - 22)°}{\text{cos}^2 22° + \text{cos}^2 (90 - 22)°}$ + sin² 63° +cos 63° sin (90 - 63)°

= $\dfrac{\text{sin}^2 22° + \text{cos}^2 22°}{\text{cos}^2 22° + \text{sin}^2 22°}$ + sin² 63° + cos 63° cos 63°

= $\dfrac{1}{1}$ + sin² 63° + cos² 63°

= 1 + 1

= 2.

Hence, the value of expression is 2.

If $\dfrac{4}{3}$ (sec² 59° - cot² 31°) - $\dfrac{2}{3}$ sin 90° + 3 tan² 56° tan² 34° = $\dfrac{x}{3}$, then find the value of x.

Answer

Solving the L.H.S of above equation using trigonometric identities,

⇒ $\dfrac{4}{3}$[sec² 59° - cot² (90 - 59)°] - $\dfrac{2}{3}$ sin 90° + 3 tan² 56° tan² (90 - 56)°

= $\dfrac{4}{3}$(sec² 59° - tan² 59°) - $\dfrac{2}{3}$ sin 90° + 3 tan² 56° cot² 56°

= $\dfrac{4}{3}$ x 1 - $\dfrac{2}{3}$ x 1 + 3 x 1

= $\dfrac{4}{3}$ - $\dfrac{2}{3}$ + 3

= $\dfrac{2}{3}$ + 3

= $\dfrac{11}{3}$.

Comparing it with R.H.S i.e.,

$\dfrac{11}{3} = \dfrac{x}{3}$

x = 11.

Hence, the value of x = 11.

Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

$\dfrac{\text{cos A}}{\text{1 - sin A}} + \dfrac{\text{cos A}}{\text{1 + sin A}} = \text{2 sec A}.$

Answer

Solving L.H.S.,

$\Rightarrow \dfrac{\text{cos A(1 + sin A) + \text{cos A(1 - sin A)}}}{\text{(1 - sin A)(1 + sin A)}} \\[1em] = \dfrac{\text{cos A + cos A sin A + cos A - cos A sin A}}{1 - \text{sin}^2 A} \\[1em] = \dfrac{2\text{ cos A}}{\text{cos}^2 A} \\[1em] = \dfrac{2}{\text{cos A}} \\[1em] = 2\text{ sec A}.$

Since, L.H.S. = R.H.S. hence, proved that $\dfrac{\text{cos A}}{\text{1 - sin A}} + \dfrac{\text{cos A}}{\text{1 + sin A}} = 2\text{ sec A}$.

Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

$\dfrac{\text{cos A}}{\text{cosec A + 1}} + \dfrac{\text{cos A}}{\text{cosec A - 1}} = \text{2 tan A}$.

Answer

Solving L.H.S.,

$\Rightarrow \dfrac{\text{cos A(cosec A - 1) + \text{cos A(cosec A + 1)}}}{\text{(cosec A - 1)(cosec A + 1)}} \\[1em] = \dfrac{\text{cos A cosec A - cos A + cos A cosec A + cos A}}{\text{cosec}^2 A - 1} \\[1em] = \dfrac{2\text{ cos A } \times \dfrac{1}{\text{sin A}}}{\text{cot}^2 A} \\[1em] = \dfrac{2\text{cot A}}{\text{cot}^2 A} \\[1em] = \dfrac{2}{\text{cot A}} \\[1em] = 2 \text{ tan A}.$

Since, L.H.S. = R.H.S. hence, proved that $\dfrac{\text{cos A}}{\text{cosec A + 1}} + \dfrac{\text{cos A}}{\text{cosec A - 1}} = 2 \text{ tan A}$.

Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

$\dfrac{\text{(cos θ - sin θ)(1 + tan θ)}}{2\text{cos}^2 \text{ θ} - 1} = \text{sec θ}.$

Answer

Solving L.H.S.,

$\Rightarrow \dfrac{\text{(cos θ - sin θ)}(1 + \dfrac{\text{sin θ}}{\text{cos θ}})}{\text{2 cos}^2 \text{ θ} - 1} \\[1em] = \dfrac{\dfrac{\text{(cos θ - sin θ)(cos θ + sin θ)}}{\text{cos θ}}}{\text{2 cos}^2 \text{ θ} - 1} \\[1em] = \dfrac{\text{cos}^2 \text{ θ} - \text{sin}^2 θ}{\text{cos θ}(\text{2 cos}^2 \text{ θ} - 1)} \\[1em] = \dfrac{\text{cos}^2 \text{ θ} - (1 - \text{cos}^2 \text{ θ})}{\text{cos θ(2 cos}^2 \text{ θ} - 1)} \\[1em] = \dfrac{\text{cos}^2 \text{ θ} + \text{cos}^2 \text{ θ} - 1}{\text{cos θ(2 cos}^2 \text{ θ} - 1)} \\[1em] = \dfrac{\text{(2 cos}^2 \text{ θ} - 1)}{\text{cos θ(2 cos}^2 \text{ θ} - 1)} \\[1em] = \dfrac{1}{\text{cos θ}} \\[1em] = \text{sec θ}.$

Since, L.H.S. = R.H.S. hence, proved that $\dfrac{\text{(cos θ - sin θ)(1 + tan θ)}}{2\text{ cos}^2 \text{ θ} - 1} = \text{sec θ}$.

Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

sin² θ + cos⁴ θ = cos² θ + sin⁴ θ.

Answer

Solving L.H.S.,

⇒ sin² θ + cos⁴ θ

= 1 - cos² θ + (cos² θ)²

= 1 - cos² θ + (1 - sin² θ)²

= 1 - cos² θ + 1 + sin⁴ θ - 2sin² θ

= 1 - cos² θ + 1 + sin⁴ θ - 2(1 - cos² θ)

= 2 - cos² θ + sin⁴ θ - 2 + 2cos² θ

= 2cos² θ - cos² θ + sin⁴ θ

= cos² θ + sin⁴ θ.

Since, L.H.S. = R.H.S. hence, proved that sin² θ + cos⁴ θ = cos² θ + sin⁴ θ.

Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

$\dfrac{\text{cot θ}}{\text{cosec θ + 1}} + \dfrac{\text{cosec θ + 1}}{\text{cot θ}} = \text{2 sec θ}.$

Answer

Solving L.H.S.,

$\Rightarrow \dfrac{\text{cot}^2 \text{ θ} + \text{(cosec θ + 1)}^2}{\text{(cot θ)(cosec θ + 1)}} \\[1em] = \dfrac{\text{cot}^2 \text{ θ} + \text{cosec}^2 \text{ θ} + 1 + \text{2 cosec θ}}{\text{(cot θ)(cosec θ + 1)}} \\[1em] = \dfrac{\text{cot}^2 \text{ θ} + 1 + \text{cosec}^2 \text{ θ} + \text{2 cosec θ}}{\text{(cot θ)(cosec θ + 1)}} \\[1em] = \dfrac{\text{cosec}^2 \text{ θ} + \text{cosec}^2 \text{ θ} + \text{2 cosec θ}}{\text{(cot θ)(cosec θ + 1)}} \\[1em] = \dfrac{2\text{ cosec}^2 \text{ θ} + \text{2 cosec θ}}{\text{(cot θ)(cosec θ + 1)}} \\[1em] = \dfrac{\text{2 cosec θ(cosec θ + 1)}}{\text{(cot θ)(cosec θ + 1)}} \\[1em] = \dfrac{\text{2 cosec θ}}{\text{cot θ}} \\[1em] = \dfrac{\dfrac{2}{\text{sin θ}}}{\dfrac{\text{cos θ}}{\text{sin θ}}} \\[1em] = \dfrac{2}{\text{cos θ}} \\[1em] = 2\text{ sec θ}.$

Since, L.H.S. = R.H.S hence proved that $\dfrac{\text{cot θ}}{\text{cosec θ + 1}} + \dfrac{\text{cosec θ + 1}}{\text{cot θ}} = 2\text{ sec θ}$.

Prove the following trigonometry identity :

(sin θ + cos θ)(cosec θ - sec θ) = cosec θ.sec θ - 2 tan θ

Answer

Solving,

$\phantom{\Rightarrow} \text{(sin θ + cos θ)(cosec θ - sec θ)} \\[1em] \Rightarrow \text{(sin θ + cos θ)} \times \Big(\dfrac{1}{\text{sin θ}} - \dfrac{1}{\text{cos θ}}\Big) \\[1em] \Rightarrow \text{(sin θ + cos θ)} \times \Big(\dfrac{\text{cos θ - sin θ}}{\text{sin θ cos θ}}\Big) \\[1em] \Rightarrow \dfrac{\text{cos}^2 θ - \text{sin}^2 θ}{\text{sin θ. cos θ}} \\[1em] \Rightarrow \dfrac{\text{1 - 2 sin}^2 \text{ θ}}{\text{sin θ.cos θ}} \quad [\because \text{cos}^2 \text{ θ} = 1 - \text{sin}^2 \text{ θ}] \\[1em] \Rightarrow \dfrac{1}{\text{sin θ.cos θ}} - \dfrac{\text{2 sin}^2 \text{ θ}}{\text{sin θ.cos θ}} \\[1em] \Rightarrow \text{cosec θ.sec θ} - \dfrac{\text{2 sin}^2 \text{ θ}}{\text{sin θ.cos θ}} \\[1em] \Rightarrow \text{cosec θ.sec θ - 2 tan θ}.$

Hence, proved that (sin θ + cos θ)(cosec θ - sec θ) = cosec θ.sec θ - 2 tan θ.

Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

sec⁴ A(1 - sin⁴ A) - 2 tan² A = 1.

Answer

Solving L.H.S.,

$\Rightarrow \dfrac{1}{\text{cos}^4 A}\text{(1 + sin }^2 A)\text{(1 - sin}^2 A) - \dfrac{2\text{sin}^2 A}{\text{cos}^2 A} \\[1em] = \dfrac{\text{(1 + sin}^2 A)\text{cos}^2 A}{{\text{cos}^4 A}} - \dfrac{2\text{sin}^2 A}{\text{cos}^2 A} \\[1em] = \dfrac{1 + \text{sin}^2 A}{\text{cos}^2 A} - \dfrac{2\text{sin}^2 A}{\text{cos}^2 A} \\[1em] = \dfrac{1 + \text{sin}^2 A - 2\text{sin}^2 A}{\text{cos}^2 A} \\[1em] = \dfrac{1 - \text{sin}^2 A}{\text{cos}^2 A} \\[1em] = \dfrac{\text{cos}^2 A}{\text{cos}^2 A} \\[1em] = 1.$

Since, L.H.S. = R.H.S. hence proved that sec⁴ A(1 - sin⁴ A) - 2tan² A = 1.

Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

$\dfrac{1}{\text{sin A + cos A + 1}} + \dfrac{1}{\text{sin A + cos A - 1}} = \text{sec A + cosec A}.$

Answer

Solving L.H.S.,

$\Rightarrow \dfrac{1}{\text{sin A + cos A + 1}} + \dfrac{1}{\text{sin A + cos A - 1}} \\[1em] = \dfrac{\text{sin A + cos A - 1 + sin A + cos A + 1}}{\text{(sin A + cos A + 1)}\text{(sin A + cos A - 1)}} \\[1em] = \dfrac{2\text{(sin A + cos A)}}{\text{(sin A + cos A)}^2 - 1} \\[1em] = \dfrac{2\text{(sin A + cos A)}}{\text{sin}^2 A + \text{cos}^2 A + \text{2 sin A cos A} - 1} \\[1em] = \dfrac{\text{2 sin A + 2 cos A}}{1 - 1 + \text{2 sin A cos A}} \\[1em] = \dfrac{\text{2 sin A + 2 cos A}}{\text{2 sin A cos A}} \\[1em] = \dfrac{\text{2 sin A}}{\text{2 sin A cos A}} + \dfrac{\text{2 cos A}}{\text{2 sin A cos A}} \\[1em] = \dfrac{1}{\text{cos A}} + \dfrac{1}{\text{sin A}} \\[1em] = \text{sec A + cosec A}.$

Since, L.H.S. = R.H.S. hence proved that $\dfrac{1}{\text{sin A + cos A + 1}} + \dfrac{1}{\text{sin A + cos A - 1}} = \text{sec A + cosec A}$.

Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

$\dfrac{\text{sin}^3 \text{ θ} + \text{cos}^3 \text{ θ}}{\text{sin θ + cos θ}} + \text{sin θ cos θ} = 1.$

Answer

Solving L.H.S.,

$\Rightarrow \dfrac{\text{sin}^3 \text{ θ} + \text{cos}^3 \text{ θ}}{\text{sin θ + cos θ}} + \text{sin θ cos θ} \\[1em] = \dfrac{\text{(sin θ + cos θ)}(\text{sin}^2 \text{ θ} + \text{cos}^2 \text{ θ} - \text{sin θ cos θ})}{\text{sin θ + cos θ}} + \text{sin θ cos θ} \\[1em] = \text{sin}^2 \text{ θ} + \text{cos}^2 \text{ θ} - \text{sin θ cos θ + sin θ cos θ} \\[1em] = 1.$

Since, L.H.S. = R.H.S. hence proved that $\dfrac{\text{sin}^3 \text{ θ} + \text{cos}^3 \text{ θ}}{\text{sin θ + cos θ}} + \text{sin θ cos θ} = 1$.

Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

(sec A - tan A)²(1 + sin A) = 1 - sin A.

Answer

Solving L.H.S.,

$\Rightarrow \Big(\dfrac{1}{\text{cos A}} - \dfrac{\text{sin A}}{\text{cos A}}\Big)^2(1 + \text{sin A}) \\[1em] = \Big(\dfrac{\text{1 - sin A}}{\text{cos A}}\Big)^2\text{(1 + sin A)} \\[1em] = \dfrac{\text{(1 - sin A)}^2\text{(1 + sin A)}}{\text{1 - sin}^2 A} \\[1em] = \dfrac{\text{(1 - sin A)}^2\text{(1 + sin A)}}{\text{(1 - sin A)(1 + sin A)}} \\[1em] = \text{1 - sin A}.$

Since, L.H.S. = R.H.S. hence proved that (sec A - tan A)²(1 + sin A) = 1 - sin A.

Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

$\dfrac{\text{cos A}}{\text{1 - tan A}} - \dfrac{\text{sin}^2 A}{\text{cos A - sin A}} = \text{sin A + cos A}.$

Answer

Solving L.H.S.,

$\Rightarrow \dfrac{\text{cos A}}{1 - \dfrac{\text{sin A}}{\text{cos A}}} - \dfrac{\text{sin}^2 A}{\text{cos A - sin A}} \\[1em] = \dfrac{\text{cos A}}{\dfrac{\text{cos A - sin A}}{\text{cos A}}} - \dfrac{\text{sin}^2 A}{\text{cos A - sin A}} \\[1em] = \dfrac{\text{cos}^2 A}{\text{cos A - sin A}} - \dfrac{\text{sin}^2 A}{\text{cos A - sin A}} \\[1em] = \dfrac{\text{cos}^2 A - \text{sin}^2 A}{\text{cos A - sin A}} \\[1em] = \dfrac{\text{(cos A - sin A)(cos A + sin A)}}{\text{(cos A - sin A)}} \\[1em] = \text{cos A + sin A}.$

Since, L.H.S. = R.H.S. hence proved that $\dfrac{\text{cos A}}{\text{1 - tan A}} - \dfrac{\text{sin}^2 A}{\text{cos A - sin A}} = \text{sin A + cos A}$.

Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

(sec A - cosec A)(1 + tan A + cot A) = tan A sec A - cot A cosec A.

Answer

Solving L.H.S.,

$\Rightarrow \text{(sec A - cosec A)(1 + tan A + cot A)} \\[1em] = \Big(\dfrac{1}{\text{cos A}} - \dfrac{1}{\text{sin A}}\Big)\Big(1 + \dfrac{\text{sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{sin A}}\Big) \\[1em] = \Big(\dfrac{\text{sin A - cos A}}{\text{sin A cos A}}\Big)\Big(\dfrac{\text{sin A cos A} + \text{sin}^2 A + \text{cos}^2 A}{\text{sin A cos A}} \Big) \\[1em] = \dfrac{(\text{sin A - cos A})(\text{sin A cos A + 1})}{\text{sin}^2 A \space \text{cos}^2 A} \\[1em]$

Now solving R.H.S.,

$\Rightarrow \text{tan A sec A - cot A cosec A} \\[1em] = \dfrac{\text{sin A}}{\text{cos A}} \times \dfrac{1}{\text{cos A}} - \dfrac{\text{cos A}}{\text{sin A}} \times \dfrac{1}{\text{sin A}} \\[1em] = \dfrac{\text{sin A}}{\text{cos}^2 A} - \dfrac{\text{cos A}}{\text{sin}^2 A} \\[1em] = \dfrac{\text{sin}^3 A - \text{cos}^3 A}{\text{sin}^2 A \text{ cos}^2 A} \\[1em] = \dfrac{\text{(sin A - cos A)(sin}^2 A + \text{cos}^2 A + \text{sin A cos A})}{\text{sin}^2 A \text{ cos}^2 A} \\[1em] = \dfrac{(\text{sin A - cos A})(\text{sin A cos A + 1})}{\text{sin}^2 A \space \text{cos}^2 A}.$

Since, L.H.S. = R.H.S. hence proved that (sec A - cosec A)(1 + tan A + cot A) = tan A sec A - cot A cosec A.

Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

$\dfrac{\text{tan}^2 \text{ θ}}{\text{tan}^2 \text{ θ} - 1} + \dfrac{\text{cosec}^2 \text{ θ}}{\text{sec}^2 \text{ θ} - \text{cosec}^2 \text{ θ}} = \dfrac{1}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}}$.

Answer

Solving L.H.S.,

$\Rightarrow \dfrac{\dfrac{\text{sin}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}}{\dfrac{\text{sin}^2 \text{ θ}}{\text{cos}^2 \text{ θ}} - 1} + \dfrac{\dfrac{1}{\text{sin}^2 \text{ θ}}}{\dfrac{1}{\text{cos}^2 \text{ θ}} - \dfrac{1}{\text{sin}^2 \text{ θ}}} \\[1em] = \dfrac{\dfrac{\text{sin}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}}{\dfrac{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}} + \dfrac{\dfrac{1}{\text{sin}^2 \text{ θ}}}{\dfrac{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}}{\text{cos}^2 \text{ θ} \text{ sin}^2 \text{ θ}}} \\[1em] = \dfrac{\text{sin}^2 \text{ θ}}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}} + \dfrac{1}{\dfrac{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}}{\text{cos}^2 \text{ θ}}} \\[1em] = \dfrac{\text{sin}^2 \text{ θ}}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}} + \dfrac{\text{cos}^2 \text{ θ}}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}} \\[1em] = \dfrac{\text{sin}^2 \text{ θ} + \text{cos}^2 \text{ θ}}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}} \\[1em] = \dfrac{1}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}}.$

Since, L.H.S. = R.H.S. hence proved that,

$\dfrac{\text{tan}^2 \text{ θ}}{\text{tan}^2 \text{ θ} - 1} + \dfrac{\text{cosec}^2 \text{ θ}}{\text{sec}^2 \text{ θ} - \text{cosec}^2 \text{ θ}} = \dfrac{1}{\text{sin}^2 \text{ θ} - \text{cos}^2 \text{ θ}}$.

$\dfrac{\text{sin A + cos A}}{\text{sin A - cos A}} + \dfrac{\text{sin A - cos A}}{\text{sin A + cos A}} = \dfrac{2}{\text{sin }^2A - \text{cos }^2A} = \dfrac{2}{1 - 2\text{cos }^2A} = \dfrac{2\text{ sec }^2A}{\text{tan }^2A - 1}$

Answer

Given,

$\Rightarrow \dfrac{\text{sin A + cos A}}{\text{sin A - cos A}} + \dfrac{\text{sin A - cos A}}{\text{sin A + cos A}}\\[1em] \Rightarrow \dfrac{\text{(sin A + cos A)}^2 + \text{(sin A - cos A)}^2}{(\text{sin} A - \text{cos} A)(\text{sin A + cos A})}\\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A + 2\text{sin A.cos A} + \text{sin}^2 A + \text{cos}^2 A - 2\text{sin A.cos A}}{\text{sin}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \dfrac{2\text{sin}^2 A + 2\text{cos}^2 A}{\text{sin}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \dfrac{2(1 - \text{cos}^2 A) + 2\text{cos}^2 A}{\text{sin}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \dfrac{2 - 2\text{cos}^2 A + 2\text{cos}^2 A}{\text{sin}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \dfrac{2}{\text{sin}^2 A - \text{cos}^2 A}$

So proved,

$\dfrac{\text{sin A + cos A}}{\text{sin A - cos A}} + \dfrac{\text{sin A - cos A}}{\text{sin A + cos A}} = \dfrac{2}{\text{sin }^2A - \text{cos }^2A}$

Now,

$\Rightarrow \dfrac{2}{\text{sin}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \dfrac{2}{(1 - \text{cos}^2 A) - \text{cos}^2 A}\\[1em] \Rightarrow \dfrac{2}{1 - \text{cos}^2 A - \text{cos}^2 A}\\[1em] \Rightarrow \dfrac{2}{1 - 2\text{cos}^2 A}$

So proved,

$\dfrac{\text{sin A + cos A}}{\text{sin A - cos A}} + \dfrac{\text{sin A - cos A}}{\text{sin A + cos A}} = \dfrac{2}{1 - 2\text{cos}^2 A}$

Now,

$\Rightarrow \dfrac{2}{\text{sin}^2 A - \text{cos}^2 A}$

Dividing numerator and denominator by cos² A, we get :

$\Rightarrow \dfrac{\dfrac{2}{\text{cos}^2 A}}{\dfrac{\text{sin}^2 A}{\text{cos}^2 A} - \dfrac{\text{cos}^2 A}{\text{cos}^2 A}}\\[1em] \Rightarrow \dfrac{2\text{sec}^2 A}{\text{tan}^2 A - 1}\\[1em]$

So proved,

$\dfrac{\text{sin A + cos A}}{\text{sin A - cos A}} + \dfrac{\text{sin A - cos A}}{\text{sin A + cos A}} = \dfrac{2\text{sec }^2A}{\text{tan }^2A - 1}$

Hence, proved that

$\dfrac{\text{sin A + cos A}}{\text{sin A - cos A}} + \dfrac{\text{sin A - cos A}}{\text{sin A + cos A}} = \dfrac{2}{\text{sin }^2A - \text{cos }^2A} = \dfrac{2}{1 - 2\text{cos }^2A} = \dfrac{2\text{sec }^2A}{\text{tan }^2A - 1}$.

Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

2(sin⁶ θ + cos⁶ θ) - 3(sin⁴ θ + cos⁴ θ) + 1 = 0.

Answer

Solving L.H.S.,

⇒ 2[(sin² θ)³ + (cos² θ)³] - 3[(sin² θ)² + (cos² θ)²] + 1

⇒ 2[(sin² θ + cos² θ)³ - 3sin² θ cos² θ(sin² θ + cos² θ)] - 3[(sin² θ + cos² θ)² - 2sin² θ cos² θ)] + 1

⇒ 2[(1)³ - 3sin² θ cos² θ(1)] - 3[(1)² - 2sin² θ cos² θ)] + 1

⇒ 2 - 6sin² θ cos² θ - 3 + 6sin² θ cos² θ + 1

⇒ 2 - 3 + 1 - 6sin² θ cos² θ + 6sin² θ cos² θ

⇒ 3 - 3

⇒ 0.

Since, L.H.S. = R.H.S. hence, proved that 2(sin⁶ θ + cos⁶ θ) - 3(sin⁴ θ + cos⁴ θ) + 1 = 0.

If cot θ + cos θ = m and cot θ - cos θ = n, then prove that (m² - n²)² = 16 mn.

Answer

Given,

cot θ + cos θ = m ....(i)
cot θ - cos θ = n ....(ii)

Adding (i) and (ii) we get,

⇒ m + n = cot θ + cos θ + cot θ - cos θ
⇒ m + n = 2 cot θ
⇒ 2 cot θ = m + n
⇒ cot θ = $\dfrac{m + n}{2}$.

∴ tan θ = $\dfrac{2}{m + n}$ ....(iii)

Subtracting (ii) from (i) we get,

m - n = cot θ + cos θ - cot θ + cos θ
m - n = 2 cos θ
cos θ = $\dfrac{m - n}{2}$.

∴ sec θ = $\dfrac{2}{m - n}$ ....(iv)

Squaring and subtracting (iii) from (iv),

$\Rightarrow \text{sec}^2 \text{ θ} - \text{tan}^2 \text{ θ} = \Big(\dfrac{2}{m - n}\Big)^2 - \Big(\dfrac{2}{m + n}\Big)^2 \\[1em] \Rightarrow 1 = \dfrac{4}{(m - n)^2} - \dfrac{4}{(m + n)^2} \\[1em] \Rightarrow 4\Big[\dfrac{1}{(m - n)^2} - \dfrac{1}{(m + n)^2}\Big] = 1 \\[1em] \Rightarrow 4\Big[\dfrac{(m + n)^2 - (m - n)^2}{(m + n)^2(m - n)^2}\Big] = 1 \\[1em] \Rightarrow 4\Big[\dfrac{m^2 + n^2 + 2mn - m^2 - n^2 + 2mn}{(m + n)(m - n)(m + n)(m - n)} \Big] = 1 \\[1em] \Rightarrow 4\Big[\dfrac{\cancel{m^2} + \cancel{n^2} + 2mn - \cancel{m^2} - \cancel{n^2} + 2mn}{(m^2 - n^2)(m^2 - n^2)} \Big] = 1 \\[1em] \Rightarrow 4 \times \dfrac{4mn}{(m^2 - n^2)^2} = 1 \\[1em] \Rightarrow 16 mn = (m^2 - n^2)^2.$

Hence, proved that (m² - n²)² = 16 mn.

When 0° < θ < 90°, solve the following equation:

2 cos² θ + sin θ - 2 = 0

Answer

Given,

2 cos² θ + sin θ - 2 = 0

On Solving,

⇒ 2(1 - sin² θ) + sin θ - 2 = 0

= 2 - 2 sin² θ + sin θ - 2 = 0

= sin θ -2 sin² θ = 0

= sin θ (1 - 2 sin θ) = 0

So, either sin θ = 0 or 1 - 2 sin θ = 0

If, sin θ = 0
sin θ = sin 0°
θ = 0°.

Given, θ > 0° hence, θ = 0° is not possible.

∴ 1 - 2 sin θ = 0

⇒ 1 = 2 sin θ

⇒ sin θ = $\dfrac{1}{2}$

⇒ sin θ = sin 30°

⇒ θ = 30°.

Hence, the value of θ = 30°.

When 0° < θ < 90°, solve the following equation:

3 cos θ = 2 sin² θ

Answer

Given,

3 cos θ = 2 sin² θ

On Solving,

⇒ 3 cos θ = 2(1 - cos² θ)

⇒ 3 cos θ = 2 - 2cos² θ

⇒ 2 cos² θ + 3 cos θ - 2 = 0

⇒ 2 cos² θ + 4 cos θ - cos θ - 2 = 0

⇒ 2 cos θ(cos θ + 2) - 1(cos θ + 2) = 0

⇒ (2 cos θ - 1)(cos θ + 2) = 0

⇒ 2 cos θ - 1 = 0 or cos θ + 2 = 0

⇒ cos θ = $\dfrac{1}{2}$ or cos θ = -2.

But cos θ = -2 is not possible.

∴ cos θ = $\dfrac{1}{2}$

⇒ cos θ = cos 60°

⇒ θ = 60°.

Hence, the value of θ = 60°.

When 0° < θ < 90°, solve the following equation:

sec² θ - 2 tan θ = 0

Answer

Given,

sec² θ - 2 tan θ = 0

On Solving,

⇒ 1 + tan² θ - 2 tan θ = 0

⇒ tan² θ - 2 tan θ + 1 = 0

⇒ (tan θ - 1)² = 0

⇒ tan θ - 1 = 0

⇒ tan θ = 1

⇒ tan θ = tan 45°

⇒ θ = 45°.

Hence, the value of θ = 45°.

When 0° < θ < 90°, solve the following equation:

tan² θ = 3 (sec θ - 1).

Answer

Given,

tan² θ = 3 (sec θ - 1)

On Solving,

⇒ sec² θ - 1 = 3 sec θ - 3

⇒ sec² θ - 1 - 3 sec θ + 3 = 0

⇒ sec² θ - 3 sec θ + 2 = 0

⇒ sec² θ - 2 sec θ - sec θ + 2 = 0

⇒ sec θ (sec θ - 2) - 1(sec θ - 2) = 0

⇒ (sec θ - 1)(sec θ - 2) = 0

⇒ sec θ - 1 = 0 or sec θ - 2 = 0

⇒ sec θ = 1 or sec θ = 2.

If, sec θ = 1
sec θ = sec 0°
θ = 0°.

Given, θ > 0° hence, θ = 0° is not possible.

∴ sec θ = 2

⇒ sec θ = sec 60°

⇒ θ = 60°.

Hence, the value of θ = 60°.