Question

if sin∅ =6/10, then tan∅ +sec∅

i h've the answer ,i know we have to solve it by Pythagoras theorem but still not getting the answer
pls help​

Answer 1

ANSWER:

• The value of tan ∅ + sec ∅ = 2

GIVEN:

• sin ∅ = 6/10.

TO FIND:

• The value of tan ∅ + sec ∅.

EXPLANATION:

$\boxed{ \bold{ \large{ \gray{Sin \ \theta = \dfrac{Opposite \ Side}{Hypotenuse}}}}}$

Let us consider a right angled triangle ABC, right angled at A and ∠B = ∅.

Hypotenuse = BC = 10

Opposite side = AC = 6

Adjacent side = AB

Pythagoras theorem:

• Square of the hyotenuse is equal to the sum of the squares of the other two sides.

BC² = AC² + AB²

10² = 6² + AB²

100 = 36 + AB²

AB² = 64

AB = 8

Adjacent side = AB = 8

$\boxed{ \bold{ \large{ \gray{Sec \ \theta = \dfrac{Hypotenuse}{Adjacent \ Side}}}}}$

Hypotenuse = BC = 10

Adjacent side = AB = 8

$\sf Sec \ \phi = \dfrac{10}{8}$

$\boxed{ \bold{ \large{ \gray{Tan \ \theta = \dfrac{Opposite\ side}{Adjacent \ Side}}}}}$

Opposite side = AC = 6

Adjacent side = AB = 8

$\sf Tan \ \phi = \dfrac{6}{8}$

$\sf Sec \ \phi + Tan \ \phi =\dfrac{10}{8} + \dfrac{6}{8}$

$\sf Sec \ \phi + Tan \ \phi =\dfrac{10 + 6}{8}$

$\sf Sec \ \phi + Tan \ \phi =\dfrac{16}{8}$

$\sf Sec \ \phi + Tan \ \phi =2$

Hence the value of tan ∅ + sec ∅ = 2

VERIFICATION:

$\sf Sec \ \phi - Tan \ \phi =\dfrac{10}{8} - \dfrac{6}{8}$

$\sf Sec \ \phi - Tan \ \phi =\dfrac{10 - 6}{8}$

$\sf Sec \ \phi - Tan \ \phi =\dfrac{4}{8}$

$\sf Sec \ \phi - Tan \ \phi =\dfrac{1}{2}$

$\boxed{ \bold{ \large{ \gray{Sec^2\ \theta - Tan^2\ \theta = 1}}}}$

(Sec ∅ + Tan ∅)(Sec ∅ - Tan ∅) = Sec² ∅ - Tan² ∅

$\sf Sec \ \phi - Tan \ \phi =\dfrac{1}{2}$

$\sf Sec \ \phi + Tan \ \phi =2$

Sec² ∅ - Tan² ∅ = 2 × 1/2

Sec² ∅ - Tan² ∅ = 1

HENCE VERIFIED.

NOTE : REFER ATTACHMENT FOR DIAGRAM.

Answer 2

Answer

2

Solution

• Given , Sin Ø = 6/10 = 3/5

Reffer attached file

• From fig , using Pythagoras theorem ,
• y²+3² = 5²
• y = 4

• Tan ø = opp/adj = 3/4
• sec ø = hyp/adj = 5/4
• Tan ø + sec ø = 3/4 + 5/4 = 8/4 = 2