Question

of tamo + seco
= 2/3
then tan O-seco:​

Answer 1

$\maltese$ Given :-

$tan\theta+sec\theta=\dfrac{2}{3}$

$\maltese$ To find :-

$tan\theta - sec\theta$

$\maltese$ SOLUTION:-

As we the trigonometric identities that is ,

$sec^2\theta- tan^2\theta = 1$

As we the algebraic identities a²-b² = (a+b)(a-b) So,

$(sec\theta+tan\theta) (sec\theta-tan\theta) = 1$

$\dfrac{2}{3} (sec\theta-tan\theta) = 1$

$sec\theta-tan\theta = \dfrac{1}{\dfrac{2}{3} }$

$sec\theta-tan\theta = \dfrac{3}{2}$

Take common " - "

$- ( -sec\theta +tan\theta) = \dfrac{3}{2}$

$tan\theta-sec\theta=\dfrac{-3}{2}$

$\maltese$Know more :-

Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trigonometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trigonometric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj

Answer 2

Question:

If $\mathsf{tan∅+sec∅=\frac{2}{3}}$ then, $\mathsf{tan∅-sec∅=?}$

Given:

$\mathsf{tan∅+sec∅=\frac{2}{3}}$

To Find:

$\mathsf{tan∅-sec∅}$

Solution:

We know,

$\mathsf{sec²∅-tan²∅=1}$

Then,

$\mathsf{(sec∅+tan∅)(sec∅-tan∅)=1}$

$\mathsf{(\frac{2}{3})(sec∅-tan∅)=1}$

$\mathsf{(sec∅-tan∅)=\frac{1}{\frac{2}{3}}}$

$\mathsf{(sec∅-tan∅)=\frac{3}{2}}$

OR,

$\mathsf{-(-sec∅+tan∅)=\frac{-3}{2}}$

$\mathsf{tan∅-sec∅=\frac{-3}{2}}$

More to know:

$\mathsf{sin∅=\large\frac{opp}{hyp}}$

$\mathsf{cos∅=\large\frac{adj}{hyp}}$

$\mathsf{tan∅=\large\frac{opp}{adj}}$

$\mathsf{cot∅ =\large\frac{adj}{opp}}$

$\mathsf{cosec∅=\large\frac{hyp}{opp}}$

$\mathsf{sec∅=\large\frac{hyp}{adj}}$

Abbreviation:

sin ≈ Sin

cos= Cosine

tan= Tangent

cot= Cotangent

cosec= Cosecant

sec= Secant