Question

if tanA= ¾ then find the value of 1/sinA + 1/cosA​

Answer 1

EXPLANATION.

⇒ tan A = 3/4.

tan ∅ = Perpendicular/base.

⇒ tan A = p/b = 3/4.

by using the Pythagorean theorem, we get

⇒ h² = p² + b².

⇒ h² = (3)² + (4)².

⇒ h² = 9 + 16.

⇒ h² = 25.

⇒ h = √25

⇒ h = 5.

sin A = perpendicular/hypotenuse = 3/5.

cos A = base/hypotenuse = 4/5.

tan A = perpendicular/base = 3/4.

cosec A = hypotenuse/perpendicular = 5/3.

sec A = hypotenuse/base = 5/4.

cot A = base/perpendicular = 4/3.

TO FIND. = 1/Sin A + 1/Cos A.

⇒ 1/3/5 + 1/4/5.

⇒ 1/1 X 5/3 + 1/1 X 5/4.

⇒ 35/12.

⇒ value of 1/sin A + 1/cos A = 35/12.

MORE INFORMATION.

(1) =  sin²A + cos²A = 1.

(2) = tan²A + 1 = sec²A.

(3) = 1 + cot²A = cosec²A.

(4) = sin 2A = 2 sinA cosA.

(5) = cos 2A = sin²A - cos²A

(6) = sin ( a ± b ) = sin a cos b ± cos a sin b.

(7) = cos ( a ± b ) = cos a cos b ± sin a sin b.

Answer 2

${\Large{\bold{\sf{\underline{Solution}}}}}$

• ${\bold{\sf{tanA = \dfrac{3}{4}}}}$

• ${\bold{\sf{\dfrac{Side \: opposite \: to \: angle \: X}{Side \: adjacent \: to \: angle \: X} = \dfrac{3}{4}}}}$

• ${\bold{\sf{\dfrac{YZ}{XY} = \dfrac{3}{4}}}}$

${\bold{\sf{Let,}}}$

• ${\bold{\sf{YZ \: = \: 3a}}}$ and ${\bold{\sf{XY \: = \: 4a}}}$

${\bold{\sf{So, \: now \: we \: have \: to \: find \: XZ \: by \: using \: phythagoras \: theorm}}}$

• (AC)² = (BC)² + (AB)²

• (AC)² = (3x)² + (4x)²

• (AC)² = (9x) + (16x)

• (AC)² = 25x

• AC = √25x

• AC = 5x

${\small{\bold{\sf{Now}}}}$

• sinA = ${\bold{\sf{\dfrac{Side \: opposite \: to \: angle \: X}{Hypotenuse}}}}$

• ${\bold{\sf{\dfrac{YZ}{XZ}}}}$

• ${\bold{\sf{\dfrac{3x}{5x}}}}$

• ${\bold{\sf{\dfrac{3}{5}}}}$

${\small{\bold{\sf{Now}}}}$

• cosA = ${\bold{\sf{\dfrac{Side \: adjacent \: to \: angle \: X}{Hypotenuse}}}}$

• ${\bold{\sf{\dfrac{XY}{XZ}}}}$

• ${\bold{\sf{\dfrac{4x}{5x}}}}$

• ${\bold{\sf{\dfrac{4}{5}}}}$

${\small{\bold{\sf{Now}}}}$

• ${\bold{\sf{\dfrac{1}{sinA}}}}$ + ${\bold{\sf{\dfrac{1}{cosA}}}}$ =

• ${\bold{\sf{\dfrac{\dfrac{1}{3}}{5}}}}$ + ${\bold{\sf{\dfrac{\dfrac{1}{4}}{5}}}}$ =

• ${\bold{\sf{\dfrac{5}{3}}}}$ + ${\bold{\sf{\dfrac{5}{4}}}}$ =

• ${\bold{\sf{\dfrac{5 \times 4 + 5 \times 3}{3 \times 4}}}}$

• ${\bold{\sf{\dfrac{20 + 15}{12}}}}$

• ${\bold{\sf{\dfrac{35}{12}}}}$

More Knowledge :

Trigonometric Identities :

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$

Trigonometric Table :

$\bullet\:\sf Trigonometric\:Values :\\\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0& \dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D{e}fined\end{tabular}}$