trigonometric ratios and identities mcq Model Questions & Answers, Practice Test for ibps rrb so officer sclae 2 3 single exam 2024

Answer: (c)

${\text"tan A - sin A"}/{sin^3 A} = {{\text"sin A"}/{\text"cos A"} - \text"sin A"}/{sin^3 A}$

Multiply in Numerator and Denominator by (1 + cos A)

= ${(\text"1 - cos A")}/{\text"cos A". sin^2 A} × {(\text"1 + cos A")}/{(\text"1 + cos A")}$

= ${(1 - cos^2 A)}/{cos A . sin^2 A (\text"1 + cos A")}$

= ${sin^2 A}/{cos A . sin^2 A (\text"1 + cos A")}$

= $1/{\text"cos A"} . 1/{\text"1 + cos A"} = {\text"sec A"}/{\text"1 + cos A"}$

Answer: (c)

Given that :

sin A = $3/5$ (A is acute, i.e. 0 ≤ A < 90°)

Then, cos A = $√{1 - sin^2 A}$

= $√{1 - (3/5)^2} = √{1 - 9/{25}}$

= $√{{16}/{25}} = 4/5$

∴ tan A + sec A = ${\text"sin A"}/{\text"cos A"} + 1/{\text"cos A"} = {\text"1 + sin A"}/{\text"cos A"}$

= ${1 + 3/5}/{4/5} = {8/5}/{4/5} = 8/4 = 2$

Answer: (d)

In ΔABC,

tan A = ${BC}/{AC} = u/v$

and tan B = $v/u$

Also, $u^2 + v^2 = w^2$ (by Pythagoras theorem).....(i)

∴ tan A + tan B = $u/v + v/u = {u^2 + v^2}/{uv} = {w^2}/{uv}$ [from equation (i)]

Answer: (d)

$x/a - y/b$ tan θ = 1 ...(1)

$x/a \text"tan θ" + y/b$ = 1 ...(2)

Multiplying (2) by tan θ and add in (1) we get

$x/a - y/b$ tan θ = 1

${x/a tan^2 θ + y/b \text"tan θ = tan θ"}/{x/a (1 + tan^2 θ) = \text"1 + tan θ"}$

⇒ $x/a = {\text"1 + tan θ"}/{1 + tan^2 θ}$

⇒ $y/b = 1 - x/a$ tan θ

= 1 - ${(\text"1 + tan θ")}/{1 + tan^2 θ}$ tan θ

$y/b = {\text"1 - tan θ"}/{1 + tan^2 θ}$

⇒ ${x^2}/{a^2} + {y^2}/{b^2} = {(\text"1 + tan θ")^2}/{(1 + tan^2 θ)^2} + {(\text"1 - tan θ")^2}/{(1 + tan^2 θ)^2}$

= ${1 + tan^2 θ + \text"2 tan θ + 1" + tan^2 θ - \text"2 tan θ"}/{(1 + tan^2 θ)^2}$

= ${2(1 + tan^2 θ)}/{(1 + tan^2 θ)^2}$

= $2/{\text"1 + tan θ"} = 2/{sec^2 θ} = 2 cos^2 θ$

∴ Option (d) is correct.

Answer: (b)

(3 sin θ + 4 cos θ) = 5

Now, squaring both sides,

9 $sin^2 θ + 16 cos^2 θ$ + 24 sin θ cos θ = 25

⇒ 9 (1 - $cos^2$ θ) + 16 (1 - $sin^2$ θ) + 24 sin θ cos θ = 25

⇒ $9 cos^2 θ + 16 sin^2 θ - 24 \text"sin θ cos θ" = 0$

⇒ $(3 cos θ - 4 sin θ)^2$ = 0

= 3 cos θ - 4 sin θ = 0