Question

isko
solve kre .......✌✌✌✌
express sin 67° +cos75° in term of trigonometric ratios of angles between 0° and 45°.​

Answer 1

Solution :

Sin67° + Cos75°

We know that,

Sin∅ = Cos( 90° - ∅ )

Cos∅ = Sin( 90° - ∅ )

Here, 67 + 23 = 90°

•°• 67 = 90 - 23

Similarly, 15 + 75 = 90°

•°• 75 = 90 - 15

=> Sin ( 90 - 23 ) + Cos ( 90 - 15 )

=> Cos23° + Sin15°

_______________________________

Some other formulas, related this :

★ Sec∅ = Cosec ( 90 - ∅ )

★ Cosec∅ = Sec ( 90 - ∅ )

★ Tan∅ = Cot ( 90 - ∅ )

★ Cot∅ = Tan ( 90 - ∅ )

Answer 2

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

In this question we will be using:-

• $\tt{sin\:\theta=cos({90}^{\circ}-\theta)}$
• $\tt{cos\:\theta=sin({90}^{\circ}-\theta)}$

$\rule{200}2$

Using these two formulas in the question,

$\leadsto\tt{sin\:67^{\circ}+cos\:75^{\circ}}$

$\leadsto\tt{sin(90^{\circ}-67^{\circ})+cos(90^{\circ}-75^{\circ}}$

$\huge\leadsto{\boxed{\tt{cos\:23^{\circ}+ sin\:15^{\circ}}}}$

$\rule{200}4$

More Formulas :-

• $\tt{sin\:\theta=cos({90}^{\circ}-\theta)}$
• $\tt{cos\:\theta=sin({90}^{\circ}-\theta)}$
• $\tt{tan\:\theta=cot({90}^{\circ}-\theta)}$
• $\tt{cot\:\theta=tan({90}^{\circ}-\theta)}$
• $\tt{sec\:\theta=cosec({90}^{\circ}-\theta)}$
• $\tt{cosec\:\theta=sec({90}^{\circ}-\theta)}$

$\rule{200}6$