Question

निम्नलिखित का मान निकालिए:
(i) sin18° / cos 72°
(ii) tan 26° / cot 64°
(iii) cos 48° - sin 42°
(iv) cosec 31° - sec 59°

Answer 1

Step-by-step explanation:

(i)

$\frac{ \sin {18}^{0} }{ \cos {72}^{0} } = \frac{ \sin( {90}^{0} - {72}^{0} ) }{ \cos {72}^{0} } = \frac{ \cos {72}^{0} }{ \cos {72}^{0} } = 1$

(ii)

$\frac{ \tan {26}^{0} }{ \cot {64}^{0} } = \frac{ \tan( {90}^{0} - {64}^{0}) }{ \cot {64}^{0} } = \frac{ \cot {64}^{0} }{ \cot {64}^{0} } = 1$

(iii) cos 48°- sin 42°

= cos 48°- sin (90°- 48°)

= cos 48°- cos 48°

= 0

(iv) cosec 51°- sec 39°

= cosec 51°- sec (90°- 51°)

= cosec 51°- cosec 51°

= 0

Answer 2

Answer with Step-by-step explanation:

(i) sin 18°/cos 72°    

= sin (90° - 18°) /cos 72°      

= cos 72° /cos 72°  

[sin (90° - θ) = cos θ]

sin 18°/cos 72° = 1

 

(ii) tan 26°/cot 64°    

= tan (90° - 36°)/cot 64°    

= cot 64°/cot 64°  

[tan (90° - θ) = cot θ]

tan 26°/cot 64° = 1

 

(iii) cos 48° - sin 42°      

= cos (90° - 42°) - sin 42°      

= sin 42° - sin 42°  

[cos (90° - θ) = sin θ]

cos 48° - sin 42° = 0

 

(iv) cosec 31° - sec 59°      

= cosec (90° - 59°) - sec 59°      

= sec 59° - sec 59°  

[cosec (90° - θ) = sec θ]

cosec 31° - sec 59°  = 0

आशा है कि यह उत्तर आपकी अवश्य मदद करेगा।।।।

इस पाठ से संबंधित कुछ और प्रश्न :  

यदि tan 2A = cot(A - 18°), जहाँ 2A एक न्यूनकोण है, तो A का मान ज्ञात कीजिए |

https://brainly.in/question/12659817

यदि tan A = cot B, तो सिद्ध कीजिए कि A + B = 90°

https://brainly.in/question/12659826