Trigonometric Ratios & Identity Model Questions Set 1 Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (a)

Given,

In ΔABC,

tan B = $k/{√3 k}$

By Pythogaros theorem,

$AB^2 + AC^2 = BC^2$

⇒ $(√3 k)^2 + (1k)^2 = BC^2$

⇒ $BC^2 = 4k^2$

⇒ BC = 2k

Answer: (a)

tan1° tan2° tan3° tan4°.......tan 89°

tan1° tan2° ......tan45°......tan 89°

= 1

Answer: (c)

Given, A = ${π}/6 \text"and B" = {π}/3$

I. L.H.S = sin A + sin B = sin ${π}/6 + sin {π}/3$

= $1/2 + {√3}/2 = {1 + √3}/2$

R.H.S = cos A + cos B = cos ${π}/6 + {cos π}/3$

${√3}/2 + 1/2 = {√3 + 1}/2$

⇒ sin A + sin B = cos A + cos B

II. L.H.S = tan A + tan B = tan ${π}/6 + tan {π}/3$

= $1/{√3} + √3 = 4/{√3}$

R.H.S = cot A + cot B = cot ${π}/6 + cot {π}/3$

= $√3 + 1/{√3} = 4/{√3}$

⇒ tan A + tan B = cot A + cot B

Both statements are true.

Alternate Method:

A + B = ${π}/6 + {π}/3 = {π}/2$

I. sin A + sin B = sin $({π}/2 - B) + sin ({π}/2 - A)$

= cos B + cos A = cos A + cos B

II. tan A + tan B = tan $({π}/2 - B) + tan ({π}/2 - A)$

= cot B + cot A = cot A + cot B

Hence, both statements are true.

Answer: (a)

1 + tan θ = $√2$

⇒ tan θ = $√2$ - 1

∴ cot θ - 1 = $1/{√2 - 1} - 1 = {√2 + 1}/{2 - 1} - 1 = √2$

Answer: (d)

Given that:

$a^2 = {\text"1 + 2 sin θ cos θ"}/{\text"1 - 2 sin θ cos θ"}$

⇒ $a^2 = {(sin^2 θ + cos^2 θ) + \text"2 sin θ . cos θ"}/{(sin^2 θ + cos^2 θ) - \text"2 sin θ . cos θ"}$

⇒ $a^2 = {(\text"sin θ + cos θ")^2}/{(\text"sin θ - cos θ")^2} ⇒ a/1 = {\text"sin θ + cos θ"}/{\text"sin θ - cos θ"}$

(applying componendo dividendo formula)

⇒ ${a + 1}/{a - 1} = {(\text"sin θ + cos θ") + (\text"sin θ - cos θ")}/{(\text"sin θ + cos θ") - (\text"sin θ - cos θ")}$

⇒ ${a + 1}/{a - 1} = {\text"2 sin θ"}/{\text"2 cos θ"}$ = tan θ