Question

plz answer the question correctly​

Answer 1

☯Given:

$\bf{\dfrac{sec15^{\circ}}{cosec75^{\circ}}+\dfrac{sin72^{\circ}}{cos18^{\circ}}-\dfrac{tan33^{\circ}}{cot57^{\circ}}}$

☯To Find:

• Value of above expression

☯Solution:

We know that,

• cosec x= sec(90°-x)

• cos x= sin(90°-x)

• cot x= tan(90°-x)

Now,

$\longrightarrow\sf{\dfrac{sec15^{\circ}}{cosec75^{\circ}}+\dfrac{sin72^{\circ}}{cos18^{\circ}}-\dfrac{tan33^{\circ}}{cot57^{\circ}}}$

$\longrightarrow\sf{\dfrac{sec15^{\circ}}{sec(90^{\circ}-75^{\circ})}+\dfrac{sin72^{\circ}}{sin(90^{\circ}-18^{\circ})}-\dfrac{tan33^{\circ}}{tan(90^{\circ}-57^{\circ})}}$

$\longrightarrow\sf{\dfrac{sec15^{\circ}}{sec15^{\circ}}+\dfrac{sin72^{\circ}}{sin72^{\circ}}-\dfrac{tan33^{\circ}}{tan33^{\circ}}}$

$\longrightarrow\sf{1+1-1}$

$\longrightarrow\sf{2-1}$

$\longrightarrow\sf\pink{1}$

Hence, the required value is 1.

Some Other Trigonometric Identities:

• sin²θ + cos²θ= 1
• 1 + tan²θ= sec²θ
• 1 + cot²θ= cosec²θ

• $\sf{sin\theta=\dfrac{1}{cosec\theta}}$

• $\sf{tan\theta=\dfrac{1}{cot\theta}}$

• $\sf{cos\theta=\dfrac{1}{sec\theta}}$

• $\sf{tan\theta=\dfrac{sin\theta}{cos\theta}}$

• $\sf{cot\theta=\dfrac{cos\theta}{sin\theta}}$

Answer 2

Answer:

1

Step-by-step explanation:

Solution:

$\frac{ \sec 15}{ \cosec75 } + \frac{ \sin 72}{ \cos18} - \frac{ \tan 33}{ \cot57 } \\ \\ \frac{ \sec (90 - 75)}{ \cosec75 } + \frac{ \sin (90 - 18)}{ \cos18} - \frac{ \tan (90 - 57)}{ \cot57 } \\ \\ \frac{ \ \cosec 75}{ \cosec75 } + \frac{ \cos 18}{ \cos18} - \frac{ \cot57}{ \cot57 } \\ \\ 1 + 1 - 1 \\ \\ 1$

Hence the required answer is 1.

Using Identity

sec(90 - ø) = cosec∅

sin(90 - ø) = cosø

tan(90 - ø) = cotø