Question

Sec 70÷cosec 20 + sec 59÷cot 31

Answer 1

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Answer 2

$\dfrac{\sec 70}{\csc 20}+\dfrac{\sec 59}{\cot 31}=1+\sec 31$

Step-by-step explanation:

Given : Expression $\frac{\sec 70}{\csc 20}+\frac{\sec 59}{\cot 31}$

To find : Solve the expression ?

Solution :

$\dfrac{\sec 70}{\csc 20}+\dfrac{\sec 59}{\cot 31}$

Re-write the value as,

$=\dfrac{\sec (90-20)}{\csc 20}+\dfrac{\sec (90-31)}{\cot 31}$

Using trigonometric values,

$\sec(90-\theta)=\csc\theta$

$=\dfrac{\csc 20}{\csc 20}+\dfrac{\csc 31}{\cot 31}$

$=1+\dfrac{1}{\sin 31}\times \dfrac{\sin 31}{\cos 31}$

$=1+\dfrac{1}{\cos 31}$

$=1+\sec 31$

Therefore, $\dfrac{\sec 70}{\csc 20}+\dfrac{\sec 59}{\cot 31}=1+\sec 31$

#Learn more

Evaluate the following :

(v)\frac{tan35^{0}}{cot55^{0}}+\frac{cot78^{0}}{tan12^{0}}-1

(vi)\frac{sec70^{0}}{cosec20^{0}}+\frac{sin59^{0}}{cos31^{0}}

(vii)cosec 31° − sec 59°

(viii)(sin 72° + cos 18°) (sin 72° − cos 18°)

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